If one zero of the polynomial $x^2 + 3x + k$ is $2$, then the value of $k$.
- (a)$-10$
- (b)$10$
- (c)$5$
- (d)$-5$
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(A) $-10$
Polynomials — Class 10 Maths PYQs
Previous-year board questions (2023–2026) with marking-scheme solutions, grouped by topic and marks.
Try each question first, then press ↓ (or tap Show Solution) to reveal the answer. Press ↓ again for the next question.
Polynomial, Degree, Value, Find k
1 Mark Questions
What should be subtracted from the polynomial $x^2 - 16x + 30$, so that $15$ is the zero of the resulting polynomial ?
- (a)$30$
- (b)$14$
- (c)$15$
- (d)$16$
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(C) $15$
What should be added from the polynomial $x^2 - 5x + 4$, so that $3$ is the zero of the resulting polynomial ?
- (a)$1$
- (b)$2$
- (c)$4$
- (d)$5$
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(B) $2$
If a polynomial $p(x)$ is given by $p(x) = x^2 - 5x + 6$, then the value of $p(1) + p(4)$ is :
- (a)$0$
- (b)$4$
- (c)$2$
- (d)$-4$
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(B) $4$
Assertion (A): Degree of a zero polynomial is not defined.
Reason (R): Degree of a non-zero constant polynomial is $0$.
Reason (R): Degree of a non-zero constant polynomial is $0$.
- (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- (c)Assertion (A) is true, but Reason (R) is false.
- (d)Assertion (A) is false, but Reason (R) is true.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
If $-4$ is a zero of the polynomial $p(x) = x^2 - x - (2 + 2k)$, then the value of $k$ is:
- (a)$3$
- (b)$9$
- (c)$6$
- (d)$-9$
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(B) $9$
Which of the following statements is true for a polynomial $p(x)$ of degree $3$?
- (a)$p(x)$ has at most two distinct zeroes.
- (b)$p(x)$ has at least two distinct zeroes.
- (c)$p(x)$ has exactly three distinct zeroes.
- (d)$p(x)$ has at most three distinct zeroes.
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(d) $p(x)$ has at most three distinct zeroes.
Assertion (A): The polynomial $p(y) = y^2 + 4y+3$ has two zeroes.
Reason (R): A quadratic polynomial can have at most two zeroes.
(a) Both Assertion (A) and Reason (R) are true, and (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Reason (R): A quadratic polynomial can have at most two zeroes.
(a) Both Assertion (A) and Reason (R) are true, and (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
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(b) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) : The polynomial $p(y) = y^2 + 4y + 3$ has two zeroes.
Reason (R) : A quadratic polynomial can have at most two zeroes.
Reason (R) : A quadratic polynomial can have at most two zeroes.
- (a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- (b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- (c)Assertion (A) is true, but Reason (R) is false.
- (d)Assertion (A) is false, but Reason (R) is true.
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(b) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
10open ↗
One of the zeroes of the polynomial $p(x) = kx^2 - 9x + 3$ is $(-\frac{3}{2})$. The value of $k$ is :
- (a)$\frac{22}{3}$
- (b)$-\frac{14}{3}$
- (c)$\frac{14}{3}$
- (d)$-\frac{22}{3}$
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(d) $-\frac{22}{3}$
2 Marks Questions
11open ↗
If the sum of the zeroes of the polynomial $p(x) = (p + 1) x^2 + (2p + 3) x + (3p + 4)$ is $-1$, then find the value of 'p'.
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Sum of zeroes $$\begin{aligned}& = -\frac{2p + 3}{p+1} = -1 \\ & p = -2\end{aligned}$$
12open ↗
If one zero of the polynomial $p(x) = (k − 1)x^2 − (4k + 1)x + 10$ is $5$, find the value of $k$.
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Since, $5$ is the zero of the given polynomial
$\therefore (k – 1) (5)^2 – (4k + 1) (5) + 10 = 0$ (1 Mark)
$\Rightarrow k = 4$ (1 Mark)
$\therefore (k – 1) (5)^2 – (4k + 1) (5) + 10 = 0$ (1 Mark)
$\Rightarrow k = 4$ (1 Mark)
4 Marks Questions
13open ↗
While playing badminton Ravi has set the barrier chain hung between two posts at the edge of the walkway of a street. It is hung in the shape of a parabola.
Based on the above information answer the following questions :
(a) Which type of the polynomial (linear, quadratic, cubic etc.) is graphically represented by a parabola ?
(b) If the polynomial represented by a parabola, intersects the $x$-axis at $-2$ and $3$ and $y$-axis at $-3$, then write the zeroes of the parabola.
(c) Find the expression for the above polynomial.
OR
(c) If the zeroes of the polynomial are $-5$ and $3$, find its expression.
Based on the above information answer the following questions :
(a) Which type of the polynomial (linear, quadratic, cubic etc.) is graphically represented by a parabola ?
(b) If the polynomial represented by a parabola, intersects the $x$-axis at $-2$ and $3$ and $y$-axis at $-3$, then write the zeroes of the parabola.
(c) Find the expression for the above polynomial.
OR
(c) If the zeroes of the polynomial are $-5$ and $3$, find its expression.
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(a) Quadratic (1 Mark)
(b) $-2$ and $3$ (1 Mark)
(c) $p(x) = k(x + 2)(x - 3)$ (1 Mark)
$p(x) = k(x^2 - x - 6)$
Using point $(0, -3)$, $-3 = k(0 - 0 - 6) \Rightarrow k = \frac{1}{2}$
$p(x) = \frac{1}{2}(x^2 - x - 6)$ (1 Mark)
OR
(c) $g(x) = (x + 5)(x - 3)$ (1 Mark)
$g(x) = x^2 + 2x - 15$ (1 Mark)
(b) $-2$ and $3$ (1 Mark)
(c) $p(x) = k(x + 2)(x - 3)$ (1 Mark)
$p(x) = k(x^2 - x - 6)$
Using point $(0, -3)$, $-3 = k(0 - 0 - 6) \Rightarrow k = \frac{1}{2}$
$p(x) = \frac{1}{2}(x^2 - x - 6)$ (1 Mark)
OR
(c) $g(x) = (x + 5)(x - 3)$ (1 Mark)
$g(x) = x^2 + 2x - 15$ (1 Mark)
Graphical Presentation of Zero
1 Mark Questions
14open ↗
The graph of $y = p(x)$ is given, for a polynomial $p(x)$. The number of zeroes of $p(x)$ from the graph is
- (a)3
- (b)1
- (c)2
- (d)0
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(B) 1
15open ↗
The graph of $y = p(x)$ is given in the adjoining figure. Zeroes of the polynomial $p(x)$ are
- (a)$-5, 7$
- (b)$-\frac{5}{2}, -\frac{7}{2}$
- (c)$-5, 0, 7$
- (d)$-5, \frac{7}{2}, 7$
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(C) $-5, 0, 7$
16open ↗
Assertion (A): If the graph of a polynomial intersects the x-axis at exactly two points, then the number of zeroes of that polynomial is $2$.
Reason (R): The number of zeroes of a polynomial is equal to the number of points where the graph of the polynomial intersects x-axis.
Reason (R): The number of zeroes of a polynomial is equal to the number of points where the graph of the polynomial intersects x-axis.
- (a)Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A).
- (b)Both Assertion (A) and Reason (R) are true. Reason (R) does not give correct explanation of (A).
- (c)Assertion (A) is true but Reason (R) is not true.
- (d)Assertion (A) is not true but Reason (R) is true.
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(A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A)
17open ↗
Assertion (A): If the graph of a polynomial touches $x$-axis at only one point, then the polynomial cannot be a quadratic polynomial.
Reason (R) : A polynomial of degree $n(n >1)$ can have at most $n$ zeroes.
Reason (R) : A polynomial of degree $n(n >1)$ can have at most $n$ zeroes.
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(d) Assertion (A) is false but Reason (R) is true.
18open ↗
The graph of a polynomial intersects the y-axis at one point and the x-axis at two points. The number of zeroes of this polynomial are :
- (a)$1$
- (b)$2$
- (c)$3$
- (d)$0$
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(B) $2$
19open ↗
If the given figure shows the graph of polynomial $y = ax^2 + bx + c$, then :
- (a)$a < 0$
- (b)$b^2 < 4ac$
- (c)$c > 0$
- (d)a and b are of same sign
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(A) $a < 0$
20open ↗
Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is:
- (a)$3$
- (b)$5$
- (c)$2$
- (d)$4$
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(C) $2$
21open ↗
The graph of $y = f(x)$ is given.
The number of zeroes of $f(x)$ is:
The number of zeroes of $f(x)$ is:
- (a)$0$
- (b)$1$
- (c)$2$
- (d)$4$
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(a) $0$
22open ↗
The graph of $y = f(x)$ is given.
The number of zeroes of $f(x)$ is :
The number of zeroes of $f(x)$ is :
- (a)$0$
- (b)$1$
- (c)$2$
- (d)$4$
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(a) $0$ (1 Mark)
23open ↗
The graph of $y = f(x)$ is given. The number of distinct zeroes of $y = f(x)$ is :
- (a)$0$
- (b)$1$
- (c)$2$
- (d)$3$
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(C) $2$ (1 Mark)
24open ↗
Observe the graph of polynomial $p(x)$. Number of zeroes of $p(x)$ is
- (a)$5$
- (b)$4$
- (c)$6$
- (d)$3$
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(D) $3$
25open ↗
Observe the graph of polynomial $p(x)$. The zeroes of the polynomial are
- (a)$-2, 0, 2.5$
- (b)$-2, 2.5$
- (c)$0,4$
- (d)$-2,0$
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(B) $-2, 2.5$
26open ↗
The graph of the polynomial $ax^2 + bx + c$ is a downward parabola if
- (a)$a > 0$
- (b)$a < 0$
- (c)$a = 0$
- (d)$a > 1$
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(B) $a < 0$
Find Zeroes
1 Mark Questions
27open ↗
The zeroes of the polynomial $3x^2 + 11x-4$ are:
- (a)$\frac{1}{2}$, $-4$
- (b)$\frac{1}{4}$, $-3$
- (c)$\frac{1}{3}$, $-4$
- (d)$\frac{1}{3}$, $4$
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(c) $\frac{1}{3}$, $-4$
28open ↗
The zeroes of the polynomial $p(x) = x^2 + 4x + 3$ are given by:
- (a)$1,3$
- (b)$-1,3$
- (c)$1,-3$
- (d)$-1,-3$
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(d) $-1, -3$
29open ↗
The zeroes of the quadratic polynomial $2x^2 - 3x - 9$ are:
- (a)$3, -\frac{3}{2}$
- (b)$-3, -\frac{3}{2}$
- (c)$-3,\frac{2}{2}$
- (d)$3,\frac{2}{2}$
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(A) $3, -\frac{3}{2}$
30open ↗
The zeroes of the quadratic polynomial $x^2+ 99x + 127$ are :
- (a)both positive
- (b)both negative
- (c)one positive and one negative
- (d)both equal
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(B) both negative
31open ↗
Zeroes of the polynomial $p(x) = x^2 -3\sqrt{2}x+4$ are:
- (a)$2,\sqrt{2}$
- (b)$2\sqrt{2},\sqrt{2}$
- (c)$4\sqrt{2},-\sqrt{2}$
- (d)$\sqrt{2},2$
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(B) $2\sqrt{2}, \sqrt{2}$
32open ↗
Zeroes of the polynomial $p(y) = 7y^2 - \frac{11}{3}y - \frac{2}{3}$ are:
- (a)$\frac{2}{3}, -\frac{1}{7}$
- (b)$-\frac{2}{7}, -\frac{1}{3}$
- (c)$\frac{2}{1}, \frac{1}{7}$
- (d)$\frac{2}{3}, -\frac{1}{7}$
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(D) $\frac{2}{3}, -\frac{1}{7}$
33open ↗
The value of $x$, for which the polynomials $9 - x^2$ and $6x + x^2 + 9$ vanish simultaneously, is
- (a)$3$
- (b)$2$
- (c)$-2$
- (d)$-3$
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(D) $-3$
34open ↗
How many zeroes does $p(x) = (x-2)(x+3)$ have?
- (a)Zero
- (b)One
- (c)Two
- (d)Three
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(C) Two
35open ↗
At which of the following points, the quadratic polynomial $p(x) = 3x + 18x^2 - 1$ intersects the positive $x$-axis ?
- (a)$(\frac{1}{6}, 0)$
- (b)$(-\frac{1}{3}, 0)$
- (c)$(-\frac{1}{6}, 0)$
- (d)$(\frac{1}{3}, 0)$
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(D) $(\frac{1}{3}, 0)$
36open ↗
The graph of a quadratic polynomial $f(x)$ passes through $(5,0)$, $(0, -1)$ and $(-2, 0)$. The two factors of the polynomial are
- (a)$(x + 2), (x - 5)$
- (b)$(x + 5), (x - 2)$
- (c)$(x + 1), (x - 5)$
- (d)$(x - 1), (x + 2)$
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(A) $(x + 2), (x - 5)$
2 Marks Questions
37open ↗
Find the zeroes of the polynomial $p(x) = x^2 + \frac{4}{3}x - \frac{4}{3}$.
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$\frac{1}{3}(3x^2 + 4x - 4) = \frac{1}{3}(3x^2 + 6x - 2x - 4) = \frac{1}{3}(3x - 2)(x + 2)$.
Zeroes are $\frac{2}{3}, -2$.
Zeroes are $\frac{2}{3}, -2$.
38open ↗
If $p$ and $q$ are zeroes of the polynomial $p(y) = 21y^2 - y - 2$, then find the value of $(1-p) . (1 - q)$.
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$p+q=\frac{1}{21}$
$p.q=\frac{-2}{21}$
$(1-p)(1 - q) = 1 - (p + q) + pq$
$= 1-\frac{1}{21} - \frac{2}{21}$
$= \frac{18}{21}$ or $\frac{6}{7}$
$p.q=\frac{-2}{21}$
$(1-p)(1 - q) = 1 - (p + q) + pq$
$= 1-\frac{1}{21} - \frac{2}{21}$
$= \frac{18}{21}$ or $\frac{6}{7}$
39open ↗
Find the zeroes of the polynomial $p(x) = 15x^2 – 19x + 6$.
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For zeroes $15x^2 – 19x + 6= 0$ (1 Mark)
Zeroes are $\frac{3}{5}$ and $\frac{2}{3}$ (1 Mark)
Zeroes are $\frac{3}{5}$ and $\frac{2}{3}$ (1 Mark)
Relationship of Zeros and Coefficients
1 Mark Questions
40open ↗
If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = 2x^2- 7x + 3$, then the value of $\alpha^2 + \beta^2$ is :
- (a)$10$
- (b)$\frac{37}{4}$
- (c)$\frac{23}{2}$
- (d)$37$
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(b) $\frac{37}{4}$
41open ↗
If the sum of the zeroes of the quadratic polynomial $p(x) = kx^2+ 2x + 3k$ is equal to the product of its zeroes, then the value of $k$ is :
- (a)$-\frac{2}{3}$
- (b)$\frac{2}{3}$
- (c)$\frac{3}{2}$
- (d)$-\frac{3}{2}$
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(a) $-\frac{2}{3}$
42open ↗
If $\alpha, \beta$ are the zeroes of a polynomial $p(x) = x^2+x-1$, then $\frac{1}{\alpha} + \frac{1}{\beta}$ equals to
- (a)1
- (b)2
- (c)-1
- (d)$-\frac{1}{2}$
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(A) 1
43open ↗
If $\alpha, \beta$ are zeros of a polynomial $P(x) = 2x^2 -x-1$ then $\alpha^2 + \beta^2$ is equal to
- (a)$-\frac{3}{4}$
- (b)$\frac{5}{4}$
- (c)$\frac{1}{4}$
- (d)$\frac{3}{4}$
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(B) $\frac{5}{4}$
44open ↗
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 - ax - b$, then the value of $\alpha^2 + \beta^2$ is :
- (a)$a^2-2b$
- (b)$a^2 + 2b$
- (c)$b^2-2a$
- (d)$b^2 + 2a$
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(b) $a^2 + 2b$
45open ↗
If one zero of the polynomial $x^2-3kx + 4k$ be twice the other, then the value of $k$ is:
- (a)$-\frac{1}{2}$
- (b)$2$
- (c)$\frac{1}{2}$
- (d)$-2$
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(b) $2$
46open ↗
If '$\alpha$' and '$\beta$' are the zeroes of the polynomial $ax^2 - 5x + c$ and $\alpha + \beta = \alpha\beta = 10$, then :
- (a)$a = 5, c = -\frac{1}{2}$
- (b)$a = 1, c = \frac{5}{2}$
- (c)$a = \frac{5}{2}, c = 1$
- (d)$a = \frac{1}{2}, c = 5$
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(d) $a = \frac{1}{2}, c = 5$
47open ↗
The sum of zeroes of the polynomial $\sqrt{2}x^2 - 17$ are given as :
- (a)$\frac{17\sqrt{2}}{2}$
- (b)$-\frac{17\sqrt{2}}{2}$
- (c)$0$
- (d)$1$
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(c) $0$
48open ↗
If $\alpha$, $\beta$ are zeroes of the polynomial $x^2-1$, then value of $(\alpha + \beta)$ is :
- (a)$2$
- (b)$1$
- (c)$-1$
- (d)$0$
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(d) $0$
49open ↗
If $\alpha$, $\beta$ are the zeroes of the polynomial $p(x) = 4x^2 – 3x – 7$, then $(\frac{1}{\alpha} + \frac{1}{\beta})$ is equal to :
- (a)$\frac{7}{3}$
- (b)$-\frac{7}{3}$
- (c)$\frac{3}{7}$
- (d)$-\frac{3}{7}$
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(d) $-\frac{3}{7}$
50open ↗
If $\alpha$ and $\beta$ are the zeroes of the polynomial $2x^2 + 5x + 1$, then the value of $\alpha + \beta + 3\alpha\beta$ is :
- (a)$-4$
- (b)$-\frac{3}{2}$
- (c)$1$
- (d)$-1$
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(D) $-1$
51open ↗
If the sum of zeroes of the polynomial $p(x) = 2x^2 - k\sqrt{2}x+1$ is $\sqrt{2}$, then value of $k$ is :
- (a)$\sqrt{2}$
- (b)$2$
- (c)$2\sqrt{2}$
- (d)$\frac{1}{2}$
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(b) $2$
52open ↗
The zeroes of a polynomial $x^2 + px + q$ are twice the zeroes of the polynomial $4x^2 - 5x - 6$. The value of $p$ is :
- (a)$-\frac{5}{2}$
- (b)$\frac{5}{2}$
- (c)$-5$
- (d)$10$
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(a) $-\frac{5}{2}$
53open ↗
If $\alpha, \beta$ are the zeroes of the polynomial $6x^2 - 5x - 4$, then $\frac{1}{\alpha} + \frac{1}{\beta}$ is equal to :
- (a)$\frac{5}{4}$
- (b)$-\frac{5}{4}$
- (c)$\frac{4}{5}$
- (d)$\frac{5}{24}$
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(b) $-\frac{5}{4}$
54open ↗
If $\alpha$ and $\beta$ are zeroes of the polynomial $5x^2 + 3x - 7$, the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is
- (a)$\frac{3}{7}$
- (b)$\frac{3}{5}$
- (c)$\frac{3}{-7}$
- (d)$\frac{5}{-7}$
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(C) $\frac{3}{7}$
55open ↗
If $\alpha$ and $\beta$ are zeroes of the polynomial $2x^2 - 9x + 5$, then value of $\alpha^2 + \beta^2$ is
- (a)$\frac{1}{4}$
- (b)$\frac{61}{4}$
- (c)$1$
- (d)$\frac{71}{4}$
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(B) $\frac{61}{4}$
56open ↗
If $\alpha$ and $\beta$ ($\alpha > \beta$) are the zeroes of the polynomial $-x^2 + 8x + 9$, then $(\alpha - \beta)$ is equal to
- (a)-10
- (b)10
- (c)$\pm 10$
- (d)8
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(B) 10
57open ↗
The ratio of the sum and product of the roots of the quadratic equation $5x^2-6x+21 = 0$ is :
- (a)$5:21$
- (b)$2:7$
- (c)$21:5$
- (d)$7:2$
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(B) $2:7$
58open ↗
If $\alpha$ and $\beta$ are zeroes of the polynomial $p(x) = kx^2 - 30x + 45k$ and $\alpha + \beta = \alpha\beta$, then the value of 'k' is :
- (a)$-\frac{2}{3}$
- (b)$-\frac{3}{2}$
- (c)$\frac{3}{2}$
- (d)$\frac{2}{3}$
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(D)$\frac{2}{3}$
59open ↗
If one zero of the quadratic polynomial $4x^2+ 4x - m$ is $\frac{3}{2}$, then the other zero is :
- (a)$\frac{2}{5}$
- (b)$\frac{5}{2}$
- (c)$-\frac{5}{2}$
- (d)$-\frac{1}{2}$
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(C)-$\frac{5}{2}$
60open ↗
If $\alpha$ and $\beta$ are the zeroes of polynomial $3x^2 + 6x + k$ such that $\alpha + \beta + \alpha\beta = -\frac{2}{3}$, then the value of $k$ is:
- (a)$-8$
- (b)$8$
- (c)$-4$
- (d)$4$
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(D) $4$
61open ↗
The sum of the zeroes of the polynomial $p(x) = 5x-7x^2 + 3$ is:
- (a)$-\frac{7}{5}$
- (b)$\frac{7}{5}$
- (c)$\frac{5}{7}$
- (d)$-\frac{5}{7}$
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(C) $\frac{5}{7}$
62open ↗
If one zero of the polynomial $q(x) = (p^2 + 4)x^2 + 65x + 4p$ is reciprocal of the other, then the value of 'p' is :
- (a)$-1$
- (b)$1$
- (c)$-2$
- (d)$2$
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(D) $2$
63open ↗
If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = x^2-ax-b$, then the value of $(\alpha + \beta + \alpha\beta)$ is equal to :
- (a)$a + b$
- (b)$-a-b$
- (c)$a-b$
- (d)$-a+b$
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(C) $a - b$
64open ↗
If the zeroes of the polynomial $ax^2+bx+\frac{2a}{b}$ are reciprocal of each other, then the value of $b$ is
- (a)2
- (b)$\frac{1}{2}$
- (c)-2
- (d)$-\frac{1}{2}$
Show SolutionHide Solution↓
(A) 2
65open ↗
If the square of the difference of the zeroes of the quadratic polynomial $y^2 + py + 36$ is equal to $81$, then the values of $p$ are
- (a)$\pm 5$
- (b)$\pm 15$
- (c)$\pm 18$
- (d)$\pm 12$
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(B) $\pm 15$
66open ↗
If $\alpha$ and $\beta$ are two zeroes of a polynomial $f(x) = px^2 - 2x + 3p$ and $\alpha + \beta = \alpha\beta$, then value of $p$ is:
- (a)$\frac{2}{3}$
- (b)$\frac{2}{3}$
- (c)$\frac{1}{3}$
- (d)$-\frac{1}{3}$
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(B) $\frac{2}{3}$
67open ↗
A polynomial $p(x)$, which has sum of its zeroes equal to their product, is :
- (a)$3x^2 + 2x + 2$
- (b)$3x^2-2x-3$
- (c)$3x^2-2x + 2$
- (d)$x^2 - 3x + 2$
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(C) $3x^2 - 2x + 2$ (1 Mark)
68open ↗
$\alpha, \beta$ are zeroes of the polynomial $2x^2 + 5x + 1$. The value of $(\frac{1}{\alpha} + \frac{1}{\beta})$ is :
- (a)$-\frac{5}{4}$
- (b)$5$
- (c)$\frac{5}{4}$
- (d)$-5$
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(d) $-5$
69open ↗
If $\alpha, \beta$ are zeroes of the polynomial $3x^2 + 14x - 5$, then the value of $3\left(\frac{\alpha + \beta}{\alpha\beta}\right)$ is :
- (a)$\frac{14}{5}$
- (b)$\frac{42}{5}$
- (c)$-\frac{14}{5}$
- (d)$-\frac{42}{5}$
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(b) $\frac{42}{5}$
70open ↗
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 - x - 4$, then the value of $(\frac{1}{\alpha} + \frac{1}{\beta})$ is
- (a)$-4$
- (b)$-\frac{1}{4}$
- (c)$\frac{1}{4}$
- (d)$4$
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(B) $-\frac{1}{4}$
71open ↗
The value of $k$ for which sum of the zeroes of the polynomial $p(x) = 3x^2 - kx + 6$ is $2$, is
- (a)$2$
- (b)$-6$
- (c)$-2$
- (d)$6$
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$6$
72open ↗
If the zeroes of the polynomial $p(x) = 2x^2- 7x + 6$ are $\alpha$ and $\beta$, then the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is
- (a)$\frac{7}{2}$
- (b)$\frac{6}{7}$
- (c)$-\frac{7}{6}$
- (d)$\frac{7}{6}$
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$\frac{7}{6}$
73open ↗
If $\alpha$ and $\beta$ are two zeroes of the quadratic polynomial $p(x) = x^2 - 11x + 30$, then $\frac{1}{\alpha} + \frac{1}{\beta}$ is equal to:
- (a)$\frac{30}{11}$
- (b)$\frac{11}{30}$
- (c)$\frac{11}{30}$
- (d)$\frac{30}{11}$
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(B) $\frac{11}{30}$
2 Marks Questions
74open ↗
If $p$ and $q$ are the zeroes of the quadratic polynomial $f(x) = 6x^2 + x - 2$, then find the value of $\frac{1}{p} + \frac{1}{q} - pq$.
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$p + q = -\frac{1}{6}$, $pq = -\frac{2}{6} = -\frac{1}{3}$
$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-\frac{1}{6}}{-\frac{1}{3}} = \frac{1}{2}$
$\frac{1}{p} + \frac{1}{q} - pq = \frac{1}{2} - (-\frac{1}{3}) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-\frac{1}{6}}{-\frac{1}{3}} = \frac{1}{2}$
$\frac{1}{p} + \frac{1}{q} - pq = \frac{1}{2} - (-\frac{1}{3}) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
75open ↗
If one zero of the polynomial $p(x) = 6x^2 + 37x – (k – 2)$ is reciprocal of the other, then find the value of $k$.
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$p(x) = 6x^2 + 37x - (k – 2)$
Let the zeroes be $\alpha, \frac{1}{\alpha}$
Product of zeroes = $\alpha . \frac{1}{\alpha} = \frac{-(k-2)}{6}$
$1 = \frac{-(k-2)}{6}$
$6 = -k + 2 \Rightarrow k = -4$
Let the zeroes be $\alpha, \frac{1}{\alpha}$
Product of zeroes = $\alpha . \frac{1}{\alpha} = \frac{-(k-2)}{6}$
$1 = \frac{-(k-2)}{6}$
$6 = -k + 2 \Rightarrow k = -4$
76open ↗
If $\alpha, \beta$ are zeroes of the polynomial $p(x) = 5x^2 - 6x + 1$, then find the value of $\alpha + \beta + \alpha\beta$.
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$\alpha + \beta = \frac{6}{5}$
$\alpha\beta = \frac{1}{5}$
$\alpha + \beta + \alpha\beta = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}$
$\alpha\beta = \frac{1}{5}$
$\alpha + \beta + \alpha\beta = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}$
77open ↗
If $\alpha$ and $\beta$ are zeroes of the quadratic polynomial $p(x) = x^2 - 5x + 4$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta$.
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$\alpha + \beta = 5$ (1/2 Mark)
$\alpha\beta = 4$ (1/2 Mark)
$\frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta = \frac{\alpha+\beta}{\alpha\beta} - 2\alpha\beta$ (1/2 Mark)
$= \frac{5}{4} - 2 \times 4 = \frac{5}{4} - 8 = \frac{5 - 32}{4} = -\frac{27}{4}$ (1/2 Mark)
$\alpha\beta = 4$ (1/2 Mark)
$\frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta = \frac{\alpha+\beta}{\alpha\beta} - 2\alpha\beta$ (1/2 Mark)
$= \frac{5}{4} - 2 \times 4 = \frac{5}{4} - 8 = \frac{5 - 32}{4} = -\frac{27}{4}$ (1/2 Mark)
78open ↗
If the zeroes of the polynomial $x^2 + ax + b$ are in the ratio $3: 4$, then prove that $12a^2 = 49b$.
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Let the zeroes are $3\alpha$ and $4\alpha$
$3\alpha + 4\alpha = -a$
$\Rightarrow 7\alpha = -a$ (1/2)
Also, $12\alpha^2 = b$ (1/2)
LHS = $12a^2 = 12 (-7\alpha)^2 = 49 \times 12(\alpha)^2 = 49b$ = RHS (1)
$3\alpha + 4\alpha = -a$
$\Rightarrow 7\alpha = -a$ (1/2)
Also, $12\alpha^2 = b$ (1/2)
LHS = $12a^2 = 12 (-7\alpha)^2 = 49 \times 12(\alpha)^2 = 49b$ = RHS (1)
79open ↗
If '$\alpha$' and '$\beta$' are the zeroes of the polynomial $p(y) = y^2 - 5y + 3$, then find the value of $\alpha^4\beta^3 + \alpha^3\beta^4$.
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$\alpha + \beta = 5$
$\alpha\beta = 3$
$\alpha^4\beta^3 + \alpha^3\beta^4 = (\alpha\beta)^3(\alpha + \beta)$
$= 27 \times 5 = 135$
$\alpha\beta = 3$
$\alpha^4\beta^3 + \alpha^3\beta^4 = (\alpha\beta)^3(\alpha + \beta)$
$= 27 \times 5 = 135$
80open ↗
If $\alpha$ and $\beta$ are zeroes of the polynomial $p(x) = x^2 - 2x - 1$, then find the value of $\frac{1}{2\alpha} + \frac{1}{2\beta} + 3\alpha\beta$.
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$$\begin{aligned}& \alpha + \beta = 2 \\ & \alpha\beta = -1 \\ & \frac{1}{2\alpha} + \frac{1}{2\beta} + 3\alpha\beta = \frac{\alpha + \beta}{2\alpha\beta} + 3\alpha\beta \\ & = \frac{2}{2(-1)} + 3(-1) = -4\end{aligned}$$
81open ↗
If $\alpha$, $\beta$ are the zeroes of the polynomial $p(x) = x^2-3x-1$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
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Here, $\alpha + \beta = 3$, $\alpha\beta = -1$ (1 Mark)
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{3}{-1}$ (1/2 Mark)
$= -3$ (1/2 Mark)
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{3}{-1}$ (1/2 Mark)
$= -3$ (1/2 Mark)
82open ↗
Find the zeroes of the quadratic polynomial $x^2+7x+10$, and verify the relationship between the zeroes and its coefficients.
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$x^2 + 7x + 10 = (x + 2)(x + 5)$ (1 Mark)
So, the zeroes of the polynomial are $-2$ and $-5$ (1/2 Mark)
Sum of zeroes $= -7 = \frac{-7}{1} = \frac{\text{coefficient of } x}{\text{coeffiecient of } x^2}$ (1/2 Mark)
Product of zeroes $= 10 = \frac{10}{1} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1/2 Mark)
So, the zeroes of the polynomial are $-2$ and $-5$ (1/2 Mark)
Sum of zeroes $= -7 = \frac{-7}{1} = \frac{\text{coefficient of } x}{\text{coeffiecient of } x^2}$ (1/2 Mark)
Product of zeroes $= 10 = \frac{10}{1} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1/2 Mark)
83open ↗
If $\alpha, \beta$ are the zeroes of the quadratic polynomial $px^2 + qx + r$, then find the value of $\alpha^3\beta + \beta^3\alpha$.
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$px^2 + qx + r$
$\alpha+\beta=-\frac{q}{p}$, $\alpha\beta = \frac{r}{p}$ (1 Mark)
$\alpha^3\beta + \beta^3\alpha$
$\quad = \alpha\beta (\alpha^2 + \beta^2)$ (1/2 Mark)
$\quad = \alpha\beta [(\alpha + \beta)^2 - 2\alpha\beta] = \frac{r}{p} [(-\frac{q}{p})^2 - 2 (\frac{r}{p})]$ (1/2 Mark)
$\quad = \frac{r}{p^3} (q^2 - 2pr)$ (1/2 Mark)
$\alpha+\beta=-\frac{q}{p}$, $\alpha\beta = \frac{r}{p}$ (1 Mark)
$\alpha^3\beta + \beta^3\alpha$
$\quad = \alpha\beta (\alpha^2 + \beta^2)$ (1/2 Mark)
$\quad = \alpha\beta [(\alpha + \beta)^2 - 2\alpha\beta] = \frac{r}{p} [(-\frac{q}{p})^2 - 2 (\frac{r}{p})]$ (1/2 Mark)
$\quad = \frac{r}{p^3} (q^2 - 2pr)$ (1/2 Mark)
84open ↗
$\alpha$ and $\beta$ are the zeroes of the polynomial $5x^2 – 16x - 10$. Find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
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$\alpha + \beta = \frac{16}{5}$, $\alpha \beta = -2$ (1 Mark)
$\therefore \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha+ \beta)^2 – 2\alpha\beta}{\alpha\beta}$ (1/2 Mark)
= $\frac{(\frac{16}{5})^2 + 4}{-2}$ (1/2 Mark)
= $\frac{\frac{256}{25} + 4}{-2} = \frac{356}{25 \times -2} = -\frac{356}{50}$ or $-\frac{178}{25}$ (1/2 Mark)
$\therefore \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha+ \beta)^2 – 2\alpha\beta}{\alpha\beta}$ (1/2 Mark)
= $\frac{(\frac{16}{5})^2 + 4}{-2}$ (1/2 Mark)
= $\frac{\frac{256}{25} + 4}{-2} = \frac{356}{25 \times -2} = -\frac{356}{50}$ or $-\frac{178}{25}$ (1/2 Mark)
85open ↗
$\alpha, \beta$ are zeroes of the polynomial $p(x) = 3x^2 - 6x - 5$. Find the value of $\frac{1}{\alpha^2} + \frac{1}{\beta^2}$.
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$\alpha + \beta = 2, \alpha \beta = -\frac{5}{3}$ (I) (1 Mark)
$\therefore \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2} = \frac{4 + \frac{10}{3}}{\frac{25}{9}}$ (II) ($\frac{1}{2}$ Mark)
$= \frac{66}{25}$ (III) ($\frac{1}{2}$ Mark)
$\therefore \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2} = \frac{4 + \frac{10}{3}}{\frac{25}{9}}$ (II) ($\frac{1}{2}$ Mark)
$= \frac{66}{25}$ (III) ($\frac{1}{2}$ Mark)
86open ↗
If one zero of the polynomial $x^2-5x-c$ is $(-1)$, find the value of $c$. Also, find the other zero.
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As $-1$ is the zero, so $1 + 5 - c = 0$ gives $c = 6$ (1 Mark)
Sum of zeroes = $5$ (1/2 Mark)
Thus, other zero is $6$ (1/2 Mark)
Sum of zeroes = $5$ (1/2 Mark)
Thus, other zero is $6$ (1/2 Mark)
87open ↗
Verify the relation between the zeroes and the coefficients of the quadratic polynomial $4x^2 – 9$.
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Zeroes of $4x^2 – 9$ are $\frac{3}{2}, -\frac{3}{2}$ (1 Mark)
Sum of zeroes = $\frac{3}{2} + (-\frac{3}{2}) = 0 = -\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}$ (1/2 Mark)
Product of zeroes = $\frac{3}{2} \times (-\frac{3}{2}) = -\frac{9}{4} = \frac{\text{Constant term}}{\text{Coefficient of x}^2}$ (1/2 Mark)
Sum of zeroes = $\frac{3}{2} + (-\frac{3}{2}) = 0 = -\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}$ (1/2 Mark)
Product of zeroes = $\frac{3}{2} \times (-\frac{3}{2}) = -\frac{9}{4} = \frac{\text{Constant term}}{\text{Coefficient of x}^2}$ (1/2 Mark)
88open ↗
Verify the relationship between the zeroes and the coefficients of the quadratic polynomial $25x^2 - 16$.
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Zeroes of $25x^2 - 16$ are $\frac{4}{5}, -\frac{4}{5}$ (1 Mark)
Sum of zeroes $= \frac{4}{5} + (-\frac{4}{5}) = 0 = \frac{-0}{25} = \frac{\text{-Coefficient of } x}{\text{Coefficient of } x^2}$ (1/2 Mark)
Product of zeroes $= \frac{4}{5} \times -\frac{4}{5} = -\frac{16}{25} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ (1/2 Mark)
Sum of zeroes $= \frac{4}{5} + (-\frac{4}{5}) = 0 = \frac{-0}{25} = \frac{\text{-Coefficient of } x}{\text{Coefficient of } x^2}$ (1/2 Mark)
Product of zeroes $= \frac{4}{5} \times -\frac{4}{5} = -\frac{16}{25} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ (1/2 Mark)
89open ↗
Verify the relationship between the zeroes and the coefficients of the quadratic polynomial $9x^2 - 25$.
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Getting zeroes: $\frac{-5}{3}$ and $\frac{5}{3}$ (1/2+1/2 Mark)
Sum of zeroes = $\frac{-5}{3} + \frac{5}{3} = 0 = \frac{-0}{9} = \frac{-\text{Coefficient of } x}{\text{Coefficient of } x^2}$ (1/2 Mark)
Product of zeroes = $\frac{-5}{3} \times \frac{5}{3} = \frac{-25}{9} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ (1/2 Mark)
Sum of zeroes = $\frac{-5}{3} + \frac{5}{3} = 0 = \frac{-0}{9} = \frac{-\text{Coefficient of } x}{\text{Coefficient of } x^2}$ (1/2 Mark)
Product of zeroes = $\frac{-5}{3} \times \frac{5}{3} = \frac{-25}{9} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ (1/2 Mark)
90open ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $p(x) = 3x^2 – 5x$.
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Zeroes are $0, \frac{5}{3}$ (1/2+1/2 Mark)
Sum of zeroes $= 0 + \frac{5}{3} = \frac{5}{3} = -\frac{\text{coefficient of x}}{\text{coefficient of x}^2}$ (1/2 Mark)
Product of zeroes$= 0 \times (\frac{5}{3}) = \frac{0}{3} = \frac{\text{constant term}}{\text{coefficient of x}^2}$ (1/2 Mark)
Sum of zeroes $= 0 + \frac{5}{3} = \frac{5}{3} = -\frac{\text{coefficient of x}}{\text{coefficient of x}^2}$ (1/2 Mark)
Product of zeroes$= 0 \times (\frac{5}{3}) = \frac{0}{3} = \frac{\text{constant term}}{\text{coefficient of x}^2}$ (1/2 Mark)
91open ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $2x^2 – 3x$.
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Zeroes are $0$ and $\frac{3}{2}$ (1/2 + 1/2 Mark)
Sum of zeroes $= 0 + \frac{3}{2} = \frac{3}{2} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1/2 Mark)
Product of zeroes $= 0 \times \frac{3}{2} = \frac{0}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1/2 Mark)
Sum of zeroes $= 0 + \frac{3}{2} = \frac{3}{2} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1/2 Mark)
Product of zeroes $= 0 \times \frac{3}{2} = \frac{0}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1/2 Mark)
92open ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $5x^2 + 2x$.
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Zeroes are $0, -\frac{2}{5}$ (1/2+1/2 Mark)
Sum of zeroes $= 0 - (\frac{2}{5}) = -\frac{2}{5} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$ (1/2 Mark)
Product of zeroes $= 0 \times (-\frac{2}{5}) = \frac{0}{5} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ (1/2 Mark)
Sum of zeroes $= 0 - (\frac{2}{5}) = -\frac{2}{5} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$ (1/2 Mark)
Product of zeroes $= 0 \times (-\frac{2}{5}) = \frac{0}{5} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ (1/2 Mark)
93open ↗
If $\alpha$, $\beta$ are the zeroes of polynomial $p(x) = 6x^2-5x - 3$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
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(a) $\alpha+\beta=\frac{5}{6}$, $\alpha\beta=-\frac{3}{6}$ (1 Mark)
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/6}{-3/6} = -\frac{5}{3}$ (1 Mark)
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/6}{-3/6} = -\frac{5}{3}$ (1 Mark)
94open ↗
One zero of the polynomial $4x^2 - 12x + (2k + 1)$ is five times the other. Find the value of $k$.
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Let zeroes of the polynomial be $\alpha$ and $5\alpha$
$\alpha + 5\alpha = \frac{12}{4} \Rightarrow \alpha = \frac{1}{2}$ (1/2 Mark)
$\alpha \times 5\alpha = \frac{2k + 1}{4}$ (1/2 Mark)
$\Rightarrow k = 2$ (1 Mark)
$\alpha + 5\alpha = \frac{12}{4} \Rightarrow \alpha = \frac{1}{2}$ (1/2 Mark)
$\alpha \times 5\alpha = \frac{2k + 1}{4}$ (1/2 Mark)
$\Rightarrow k = 2$ (1 Mark)
95open ↗
If $\alpha$, $\beta$ are the zeroes of polynomial $p(x) = -9x^2 - 6x + 1$. Find the value of $\alpha^2 + \beta^2$.
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$\alpha + \beta = -\frac{-6}{9}$, $\alpha\beta = -\frac{1}{9}$ (1)
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$= \frac{2}{3}$ (1)
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$= \frac{2}{3}$ (1)
96open ↗
If $\alpha$, $\beta$ are zeroes of the polynomial $x^2 - 6x + 7$, then find the value of $4(\frac{1}{\alpha^2} + \frac{1}{\beta^2})$.
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$\alpha + \beta = 6$ and $\alpha\beta = 7$ (1/2 Mark)
$4(\frac{1}{\alpha^2} + \frac{1}{\beta^2}) = 4(\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2})$
$= 4(\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha^2\beta^2})$
$= 4(\frac{36-14}{49})$ (1/2 Mark)
$= 4(\frac{22}{49})$
$= \frac{88}{49}$
$4(\frac{1}{\alpha^2} + \frac{1}{\beta^2}) = 4(\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2})$
$= 4(\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha^2\beta^2})$
$= 4(\frac{36-14}{49})$ (1/2 Mark)
$= 4(\frac{22}{49})$
$= \frac{88}{49}$
3 Marks Questions
97open ↗
If $\alpha$ and $\beta$ are roots of the quadratic equation $x^2 - 7x + 10 = 0$, find the quadratic equation whose roots are $\alpha^2$ and $\beta^2$.
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$x^2 - 7x + 10 = 0$
$\alpha + \beta = 7, \alpha\beta = 10$
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 49 - 20 = 29$
$\alpha^2\beta^2 = (10)^2 = 100$
Quadratic Equation with roots $\alpha^2, \beta^2$ is
$\therefore x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0$
i.e. $x^2 - 29x + 100 = 0$
$\alpha + \beta = 7, \alpha\beta = 10$
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 49 - 20 = 29$
$\alpha^2\beta^2 = (10)^2 = 100$
Quadratic Equation with roots $\alpha^2, \beta^2$ is
$\therefore x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0$
i.e. $x^2 - 29x + 100 = 0$
98open ↗
If one root of the quadratic equation $x^2 + 12x - k = 0$ is thrice the other root, then find the value of $k$.
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$x^2 + 12x - k = 0$
Let the roots be $\alpha, 3\alpha$
$\alpha + 3\alpha = -12 \Rightarrow \alpha =-3$
$\alpha.3\alpha=-k \Rightarrow 3\alpha^2 = -k$
$\Rightarrow k = -27$
Let the roots be $\alpha, 3\alpha$
$\alpha + 3\alpha = -12 \Rightarrow \alpha =-3$
$\alpha.3\alpha=-k \Rightarrow 3\alpha^2 = -k$
$\Rightarrow k = -27$
99open ↗
If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = x^2 – (k + 5)x + (5k + 1)$ such that, $\alpha + \beta = \frac{\alpha\beta}{3}$, then find the value of k.
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Here, $\alpha + \beta = (k + 5)$ and $\alpha\beta = (5k + 1)$
Given, $\alpha + \beta = \frac{\alpha\beta}{3}$
$k+5 = \frac{5k+1}{3}$
$\Rightarrow k = 7$
Given, $\alpha + \beta = \frac{\alpha\beta}{3}$
$k+5 = \frac{5k+1}{3}$
$\Rightarrow k = 7$
100open ↗
If $\alpha, \beta$ are the zeroes of the polynomial $3x^2 - 13x - 10$, then find the value of $(3\alpha + 1) (3\beta + 1)$.
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$\alpha, \beta$ are zeroes of $3x^2 - 13x - 10$
$\therefore \alpha + \beta = \frac{13}{3}$, $\alpha\beta = \frac{-10}{3}$
$(3\alpha + 1). (3\beta + 1) = 9\alpha\beta + 3(\alpha + \beta) + 1$
$= -30 + 13 + 1$
$= -16$
$\therefore \alpha + \beta = \frac{13}{3}$, $\alpha\beta = \frac{-10}{3}$
$(3\alpha + 1). (3\beta + 1) = 9\alpha\beta + 3(\alpha + \beta) + 1$
$= -30 + 13 + 1$
$= -16$
101open ↗
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = 6x^2 + 11x - 10$, find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
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Here $\alpha+\beta = -\frac{11}{6}$ and $$\begin{aligned}& \alpha\beta = -\frac{10}{6} \\ & \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} \\ & = \frac{(-\frac{11}{6})^2-2\times(-\frac{10}{6})}{-\frac{10}{6}} \\ & = \frac{\frac{121}{36}+\frac{20}{6}}{-\frac{10}{6}} = \frac{\frac{121+120}{36}}{-\frac{10}{6}} = \frac{241}{36} \times -\frac{6}{10} = -\frac{241}{60}\end{aligned}$$
102open ↗
Find the zeroes of the polynomial $2t^2 - 9t - 45$ and verify the relationship between the zeroes and the coefficients of the polynomial.
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$2t^2 - 9t - 45 = 2t^2 - 15t + 6t - 45$
$= (2t - 15) (t + 3)$
$\therefore$ zeroes of the polynomial are $\frac{15}{2}$ and $-3$.
Sum of the zeroes = $\frac{15}{2} + (-3) = \frac{9}{2} = -\frac{\text{coefficient of } t}{\text{coefficient of } t^2}$
Product of the zeroes = $\frac{15}{2} \times (-3) = -\frac{45}{2} = \frac{\text{constant term}}{\text{coefficient of } t^2}$
$= (2t - 15) (t + 3)$
$\therefore$ zeroes of the polynomial are $\frac{15}{2}$ and $-3$.
Sum of the zeroes = $\frac{15}{2} + (-3) = \frac{9}{2} = -\frac{\text{coefficient of } t}{\text{coefficient of } t^2}$
Product of the zeroes = $\frac{15}{2} \times (-3) = -\frac{45}{2} = \frac{\text{constant term}}{\text{coefficient of } t^2}$
103open ↗
Find the zeroes of the quadratic polynomial $x^2 - 15$ and verify the relationship between the zeroes and the coefficients of the polynomial.
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Let $P(x) = x^2 - 15$
$= (x - \sqrt{15})(x + \sqrt{15})$
$\therefore$ Zeroes of $P(x)$ are $-\sqrt{15}$ and $\sqrt{15}$
Verification-
Sum of zeroes = $-\sqrt{15} + \sqrt{15} = \frac{0}{1} = \frac{- \text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes = $-\sqrt{15} \times \sqrt{15} = -15 = \frac{-15}{1} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
$= (x - \sqrt{15})(x + \sqrt{15})$
$\therefore$ Zeroes of $P(x)$ are $-\sqrt{15}$ and $\sqrt{15}$
Verification-
Sum of zeroes = $-\sqrt{15} + \sqrt{15} = \frac{0}{1} = \frac{- \text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes = $-\sqrt{15} \times \sqrt{15} = -15 = \frac{-15}{1} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
104open ↗
Find the zeroes of the polynomial $4x^2 + 4x - 3$ and verify the relationship between zeroes and coefficients of the polynomial.
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$$\begin{aligned}& P(x) = 4x^2 + 4x - 3 \\ & = (2x + 3) (2x - 1) \\ & \therefore \text{Zeroes of the polynomial are } -\frac{3}{2}, \frac{1}{2} \\ & \text{Sum of Zeroes } = -\frac{3}{2} + \frac{1}{2} = \frac{-3+1}{2} = \frac{-4}{4} = -1 = -\frac{(\text{coefficient of } x)}{(\text{coefficient of } x^2)} \\ & \text{Product of Zeroes } = -\frac{3}{2} \times \frac{1}{2} = -\frac{3}{4} = \frac{\text{constant term}}{(\text{coefficient of } x^2)}\end{aligned}$$
105open ↗
If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 + x - 2$, then find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
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Here $\alpha + \beta = - 1$ and $$\begin{aligned}& \alpha\beta = -2 \\ & \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} \\ & = \frac{(-1)^2-2(-2)}{-2} = \frac{1+4}{-2} = -\frac{5}{2}\end{aligned}$$
106open ↗
Find the zeroes of the polynomial $f(t) = t^2 + 4\sqrt{3}t - 15$ and verify the relationship between the zeroes and the coefficients of the polynomial.
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$$\begin{aligned}& f(t) = t^2 + 4\sqrt{3}t - 15 \\ & = t^2 + 5\sqrt{3}t - \sqrt{3}t - 15 \\ & = (t - \sqrt{3}) (t + 5\sqrt{3})\ \therefore\end{aligned}$$ Zeroes of given polynomial are $$\begin{aligned}& -5\sqrt{3}, \sqrt{3} \\ & Sum of the zeroes =\end{aligned}$$(-5√3 + √3) = -4√3 = -coefficient of t/coefficient of t²
Product of the zeroes = $(-5\sqrt{3}) \times \sqrt{3}) = -15 = \frac{\text{constant term}}{\text{coefficient of t}^2}$
Product of the zeroes = $(-5\sqrt{3}) \times \sqrt{3}) = -15 = \frac{\text{constant term}}{\text{coefficient of t}^2}$
107open ↗
$\alpha, \beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 - 4x + k$, such that $\alpha - \beta = 8$. Find the value of $k$.
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$p(x) = x^2 - 4x + k$
Here, $\alpha + \beta = 4$
and $\alpha \beta = k$
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$
$\Rightarrow (8)^2 = (4)^2 - 4k$
$\Rightarrow k = -12$
Here, $\alpha + \beta = 4$
and $\alpha \beta = k$
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$
$\Rightarrow (8)^2 = (4)^2 - 4k$
$\Rightarrow k = -12$
108open ↗
Zeroes of the quadratic polynomial $p(x) = (a^2 + 10)x^2 - 74x + 7a$ are reciprocal of each other and they are rational. Find the value of 'a'.
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Since zeroes are reciprocal of each other.
$\frac{7a}{a^2+10} = 1$
$\Rightarrow a^2 - 7a + 10 = 0$
$\Rightarrow (a-2)(a-5) = 0$
$\Rightarrow a = 2, 5$
For $a = 5$, zeroes are rational.
$\frac{7a}{a^2+10} = 1$
$\Rightarrow a^2 - 7a + 10 = 0$
$\Rightarrow (a-2)(a-5) = 0$
$\Rightarrow a = 2, 5$
For $a = 5$, zeroes are rational.
109open ↗
If $\alpha, \beta$ are the zeroes of the polynomial $p(x) = x^2 - 2x - 3$, then find a polynomial where zeroes are $(2\alpha + 3\beta)$ and $(3\alpha + 2\beta)$.
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$p(x) = x^2-2x-3$
Here $\alpha + \beta = 2$ and $\alpha\beta = - 3$
Let the required polynomial be $x^2 - Sx + P$
$S = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5(\alpha + \beta)$
$= 5 \times 2 = 10$
$P = (2\alpha + 3\beta) \times (3\alpha + 2\beta)$
$= 6 (\alpha^2 + \beta^2) + 13 \alpha\beta$
$= 6 [(\alpha + \beta)^2 - 2 \alpha\beta] + 13 \alpha\beta$
$= 6 (\alpha + \beta)^2 + \alpha\beta$
$= 6 (2)^2 + (-3)$
$= 21$
So, required polynomial is $x^2 - 10x + 21$
Here $\alpha + \beta = 2$ and $\alpha\beta = - 3$
Let the required polynomial be $x^2 - Sx + P$
$S = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5(\alpha + \beta)$
$= 5 \times 2 = 10$
$P = (2\alpha + 3\beta) \times (3\alpha + 2\beta)$
$= 6 (\alpha^2 + \beta^2) + 13 \alpha\beta$
$= 6 [(\alpha + \beta)^2 - 2 \alpha\beta] + 13 \alpha\beta$
$= 6 (\alpha + \beta)^2 + \alpha\beta$
$= 6 (2)^2 + (-3)$
$= 21$
So, required polynomial is $x^2 - 10x + 21$
110open ↗
$\alpha$ and $\beta$ are zeroes of a quadratic polynomial $px^2+qx+1$. Form a quadratic polynomial whose zeroes are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$.
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$\alpha + \beta = -\frac{q}{p}, \alpha\beta = \frac{1}{p}$
Sum of zeroes of the required polynomial = $\frac{2}{\alpha} + \frac{2}{\beta} = 2\frac{(\beta+\alpha)}{\alpha\beta} = -2q$
Product of zeroes of the required polynomial = $\frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha\beta} = 4p$
$\therefore$ required polynomial is $x^2 + 2qx + 4p$
Sum of zeroes of the required polynomial = $\frac{2}{\alpha} + \frac{2}{\beta} = 2\frac{(\beta+\alpha)}{\alpha\beta} = -2q$
Product of zeroes of the required polynomial = $\frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha\beta} = 4p$
$\therefore$ required polynomial is $x^2 + 2qx + 4p$
111open ↗
$\alpha$ and $\beta$ are zeroes of a quadratic polynomial $x^2 - ax - b$. Obtain a quadratic polynomial whose zeroes are $3\alpha + 1$ and $3\beta + 1$.
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$\alpha + \beta = a, \alpha\beta = -b$
Sum of zeroes of required polynomial
$= (3\alpha + 1) + (3\beta + 1) = 3(\alpha + \beta) + 2 = 3a + 2$
Product of zeroes of required polynomial
$= (3\alpha + 1)(3\beta + 1) = 9\alpha\beta + 3(\alpha + \beta) + 1 = -9b + 3a + 1$
$\therefore$ The required polynomial is $x^2 - (3a + 2)x + (3a - 9b + 1)$
Sum of zeroes of required polynomial
$= (3\alpha + 1) + (3\beta + 1) = 3(\alpha + \beta) + 2 = 3a + 2$
Product of zeroes of required polynomial
$= (3\alpha + 1)(3\beta + 1) = 9\alpha\beta + 3(\alpha + \beta) + 1 = -9b + 3a + 1$
$\therefore$ The required polynomial is $x^2 - (3a + 2)x + (3a - 9b + 1)$
112open ↗
If $\alpha$ and $\beta$ are the zeroes of the polynomial $ax^2 - x + c$. Obtain a polynomial whose zeroes are $\alpha - 3$ and $\beta - 3$.
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$\alpha + \beta = \frac{1}{a}, \alpha\beta = \frac{c}{a}$.
Sum of zeroes of required polynomial = $\alpha + \beta - 6 = \frac{1}{a} - 6$ or $\frac{1-6a}{a}$.
Product of zeroes of required polynomial = $\alpha\beta - 3(\alpha + \beta) + 9 = \frac{c}{a} - \frac{3}{a} + 9$.
$\therefore$ required polynomial is $x^2 - (\frac{1-6a}{a})x + \frac{c-3+9a}{a}$ or $ax^2 - (1 - 6a)x + (c - 3 + 9a)$.
Sum of zeroes of required polynomial = $\alpha + \beta - 6 = \frac{1}{a} - 6$ or $\frac{1-6a}{a}$.
Product of zeroes of required polynomial = $\alpha\beta - 3(\alpha + \beta) + 9 = \frac{c}{a} - \frac{3}{a} + 9$.
$\therefore$ required polynomial is $x^2 - (\frac{1-6a}{a})x + \frac{c-3+9a}{a}$ or $ax^2 - (1 - 6a)x + (c - 3 + 9a)$.
113open ↗
Obtain the zeroes of the polynomial $7x^2 + 18x - 9$. Hence, write a polynomial each of whose zeroes is twice the zeroes of given polynomial.
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$7x^2 + 18x - 9 = (7x - 3)(x + 3)$
$\therefore$ Zeroes are $-3, \frac{3}{7}$
New zeroes are $-6, \frac{6}{7}$
Sum of new zeroes $= (-6) + \frac{6}{7} = -\frac{36}{7}$
Product of new zeroes $= (-6) \times \frac{6}{7} = -\frac{36}{7}$
$\therefore$ Required polynomial is $x^2 + \frac{36}{7}x - \frac{36}{7}$ or $7x^2 + 36x - 36$
$\therefore$ Zeroes are $-3, \frac{3}{7}$
New zeroes are $-6, \frac{6}{7}$
Sum of new zeroes $= (-6) + \frac{6}{7} = -\frac{36}{7}$
Product of new zeroes $= (-6) \times \frac{6}{7} = -\frac{36}{7}$
$\therefore$ Required polynomial is $x^2 + \frac{36}{7}x - \frac{36}{7}$ or $7x^2 + 36x - 36$
114open ↗
Obtain the zeroes of the polynomial $p(x) = 2x^2 - 5x - 3$. Hence, obtain a polynomial each of whose zeroes is one less than each of the zero of $p(x)$.
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$p(x) = 2x^2 - 5x - 3 = (x - 3)(2x + 1)$
$\therefore$ Zeroes are $3, -\frac{1}{2}$
New zeroes are $2, -\frac{3}{2}$
Sum of new zeroes $= 2 + (-\frac{3}{2}) = \frac{1}{2}$
Product of new zeroes $= 2 \times (-\frac{3}{2}) = -3$
$\therefore$ Required polynomial is $x^2 - \frac{1}{2}x - 3$ or $2x^2 - x - 6$
$\therefore$ Zeroes are $3, -\frac{1}{2}$
New zeroes are $2, -\frac{3}{2}$
Sum of new zeroes $= 2 + (-\frac{3}{2}) = \frac{1}{2}$
Product of new zeroes $= 2 \times (-\frac{3}{2}) = -3$
$\therefore$ Required polynomial is $x^2 - \frac{1}{2}x - 3$ or $2x^2 - x - 6$
115open ↗
Find the zeroes of the polynomial $p(x) = 6x^2 - 5x - 1$. Hence, obtain a polynomial each of whose zeroes is three times the zeroes of $p(x)$.
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$p(x) = 6x^2 - 5x - 1 = (x - 1)(6x + 1)$
$\therefore$ Zeroes are $1, -\frac{1}{6}$
New zeroes are $3, -\frac{1}{2}$
Sum of new zeroes = $3 + (-\frac{1}{2}) = \frac{5}{2}$
Product of new zeroes = $3 \times (-\frac{1}{2}) = -\frac{3}{2}$
$\therefore$ Required polynomial is $x^2 - \frac{5}{2}x - \frac{3}{2}$ or $2x^2 - 5x - 3$
$\therefore$ Zeroes are $1, -\frac{1}{6}$
New zeroes are $3, -\frac{1}{2}$
Sum of new zeroes = $3 + (-\frac{1}{2}) = \frac{5}{2}$
Product of new zeroes = $3 \times (-\frac{1}{2}) = -\frac{3}{2}$
$\therefore$ Required polynomial is $x^2 - \frac{5}{2}x - \frac{3}{2}$ or $2x^2 - 5x - 3$
116open ↗
Find the zeroes of the polynomial $p(x) = 3x^2 - 4x - 4$. Hence, write a polynomial whose each of the zeroes is 2 more than zeroes of $p(x)$.
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$p(x) = 3x^2 - 4x - 4$.
Zeroes are $-\frac{2}{3}$ and 2 (1 mark).
New zeroes are $\frac{4}{3}$ and 4 ($\frac{1}{2}$ mark).
Sum of new zeroes = $\frac{4}{3} + 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark).
Product of new zeroes = $\frac{4}{3} \times 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark).
Required polynomial is $x^2 - \frac{16x}{3} + \frac{16}{3}$ or $3x^2 - 16x + 16$ ($\frac{1}{2}$ mark).
Zeroes are $-\frac{2}{3}$ and 2 (1 mark).
New zeroes are $\frac{4}{3}$ and 4 ($\frac{1}{2}$ mark).
Sum of new zeroes = $\frac{4}{3} + 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark).
Product of new zeroes = $\frac{4}{3} \times 4 = \frac{16}{3}$ ($\frac{1}{2}$ mark).
Required polynomial is $x^2 - \frac{16x}{3} + \frac{16}{3}$ or $3x^2 - 16x + 16$ ($\frac{1}{2}$ mark).
117open ↗
Find the zeroes of the polynomial $q(x) = 8x^2 - 2x - 3$. Hence, find a polynomial whose zeroes are $2$ less than the zeroes of $q(x)$.
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$p(x) = 8x^2 - 2x - 3$.
Zeroes are $-\frac{1}{2}$ and $\frac{3}{4}$.
New zeroes are $-\frac{5}{2}$ and $-\frac{5}{4}$.
Sum of new zeroes = $-\frac{5}{2} + (-\frac{5}{4}) = -\frac{15}{4}$.
Product of new zeroes = $(-\frac{5}{2}) \times (-\frac{5}{4}) = \frac{25}{8}$.
Required polynomial is $x^2 + \frac{15}{4}x + \frac{25}{8}$ or $8x^2 + 30x + 25$.
Zeroes are $-\frac{1}{2}$ and $\frac{3}{4}$.
New zeroes are $-\frac{5}{2}$ and $-\frac{5}{4}$.
Sum of new zeroes = $-\frac{5}{2} + (-\frac{5}{4}) = -\frac{15}{4}$.
Product of new zeroes = $(-\frac{5}{2}) \times (-\frac{5}{4}) = \frac{25}{8}$.
Required polynomial is $x^2 + \frac{15}{4}x + \frac{25}{8}$ or $8x^2 + 30x + 25$.
118open ↗
Find the zeroes of the polynomial $r(x) = 4x^2 + 3x - 1$. Hence, write a polynomial whose zeroes are reciprocal of the zeroes of polynomial $r(x)$.
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$p(x) = 4x^2 + 3x - 1$. Zeroes are $\frac{1}{4}, -1$.
New zeroes $4, -1$.
Sum of new zeroes $= 4 + (-1) = 3$.
Product of zeroes $= 4 \times (-1) = -4$.
Required polynomial is $(x^2 - 3x - 4)$
New zeroes $4, -1$.
Sum of new zeroes $= 4 + (-1) = 3$.
Product of zeroes $= 4 \times (-1) = -4$.
Required polynomial is $(x^2 - 3x - 4)$
119open ↗
Find the value of $p$, for which one zero of the quadratic polynomial $px^2-14x + 8$ is $6$ times the other.
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Let the zeroes be $\alpha$ and $6\alpha$
Sum of zeroes $= \alpha + 6\alpha = \frac{14}{p}$ (1/2 Mark)
⇒ 7α = $\frac{14}{p}$ ⇒ α = $\frac{2}{p}$ (1/2 Mark)
Product of zeroes $= \alpha \times 6\alpha = $8/p (1/2 Mark)
⇒ 6(2/p)² = 8/p (1/2 Mark)
⇒ p = 3 (1 Mark)
Sum of zeroes $= \alpha + 6\alpha = \frac{14}{p}$ (1/2 Mark)
⇒ 7α = $\frac{14}{p}$ ⇒ α = $\frac{2}{p}$ (1/2 Mark)
Product of zeroes $= \alpha \times 6\alpha = $8/p (1/2 Mark)
⇒ 6(2/p)² = 8/p (1/2 Mark)
⇒ p = 3 (1 Mark)
120open ↗
Find the zeroes of the polynomial $p(x) = 3x^2 - 2x - 1$ and verify the relationship between the zeroes of $p(x)$ and the coefficients of $p(x)$.
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$p(x) = 3x^2 - 2x - 1$
$= (3x + 1)(x - 1)$
Zeroes are $x = -\frac{1}{3}, 1$ [1 mark]
$\text{Sum of zeroes} = -\frac{1}{3} + 1 = \frac{2}{3} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$ [1 mark]
$\text{Product of zeroes} = -\frac{1}{3} \times 1 = \frac{-1}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ [1 mark]
$= (3x + 1)(x - 1)$
Zeroes are $x = -\frac{1}{3}, 1$ [1 mark]
$\text{Sum of zeroes} = -\frac{1}{3} + 1 = \frac{2}{3} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$ [1 mark]
$\text{Product of zeroes} = -\frac{1}{3} \times 1 = \frac{-1}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ [1 mark]
121open ↗
Find the zeroes of the polynomial $p(x) = 2x^2 + 5x + 2$ and verify the relationship between zeroes of $p(x)$ and its coefficients.
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$p(x) = 2x^2 + 5x + 2 = (2x + 1)(x + 2)$
Zeroes are $-\frac{1}{2}, -2$
Sum of zeroes $= -\frac{1}{2} + (-2) = -\frac{5}{2} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= -\frac{1}{2} \times (-2) = \frac{2}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
Zeroes are $-\frac{1}{2}, -2$
Sum of zeroes $= -\frac{1}{2} + (-2) = -\frac{5}{2} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= -\frac{1}{2} \times (-2) = \frac{2}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
122open ↗
Find the zeroes of the polynomial $q(x) = 6x^2 - 5x - 1$ and verify the relationship between the zeroes of $q(x)$ and its coefficients.
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$q(x) = 6x^2 - 5x - 1$
$(6x + 1) (x - 1) = 0$
Zeroes are $x = -\frac{1}{6}, 1$ [1 mark]
Sum of zeroes $= -\frac{1}{6} + 1 = \frac{5}{6} = \frac{- \text{Coefficient of } x}{\text{Coefficient of } x^2}$ [1 mark]
Product of zeroes $= -\frac{1}{6} \times 1 = -\frac{1}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ [1 mark]
$(6x + 1) (x - 1) = 0$
Zeroes are $x = -\frac{1}{6}, 1$ [1 mark]
Sum of zeroes $= -\frac{1}{6} + 1 = \frac{5}{6} = \frac{- \text{Coefficient of } x}{\text{Coefficient of } x^2}$ [1 mark]
Product of zeroes $= -\frac{1}{6} \times 1 = -\frac{1}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$ [1 mark]
123open ↗
Find the zeroes of the polynomial $9s^2 - 6s + 1$ and verify the relationship between the zeroes and the coefficients of the given polynomial.
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$9s^2 - 6s + 1 = (3s - 1)(3s - 1)$.
Zeroes are $\frac{1}{3}$ and $\frac{1}{3}$.
Sum of zeroes $= \frac{1}{3} + \frac{1}{3} = \frac{2}{3} = \frac{-(-6)}{9} = \frac{-\text{Coefficient of } s}{\text{Coefficient of } s^2}$.
Product of zeroes $= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} = \frac{\text{Constant term}}{\text{Coefficient of } s^2}$. ($1 + 1 + 1$ marks)
Zeroes are $\frac{1}{3}$ and $\frac{1}{3}$.
Sum of zeroes $= \frac{1}{3} + \frac{1}{3} = \frac{2}{3} = \frac{-(-6)}{9} = \frac{-\text{Coefficient of } s}{\text{Coefficient of } s^2}$.
Product of zeroes $= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} = \frac{\text{Constant term}}{\text{Coefficient of } s^2}$. ($1 + 1 + 1$ marks)
124open ↗
Find the zeroes of the polynomial $4x^2 + 4x + 1$ and verify the relationship between the zeroes and the coefficients of the given polynomial.
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$4x^2 + 4x + 1$
$(2x + 1)(2x + 1)$
Zeroes are $-\frac{1}{2}$ and $-\frac{1}{2}$ [$1$ mark]
Sum of zeroes $= -\frac{1}{2} + (-\frac{1}{2}) = -1 = \frac{-4}{4} = \frac{-\text{Coefficient of } x}{\text{Coefficient of } x^2}$ [$1$ mark]
Product of zeroes $= -\frac{1}{2} \times -\frac{1}{2} = \frac{1}{4} = \frac{\text{constant term}}{\text{Coefficient of } x^2}$ [$1$ mark]
$(2x + 1)(2x + 1)$
Zeroes are $-\frac{1}{2}$ and $-\frac{1}{2}$ [$1$ mark]
Sum of zeroes $= -\frac{1}{2} + (-\frac{1}{2}) = -1 = \frac{-4}{4} = \frac{-\text{Coefficient of } x}{\text{Coefficient of } x^2}$ [$1$ mark]
Product of zeroes $= -\frac{1}{2} \times -\frac{1}{2} = \frac{1}{4} = \frac{\text{constant term}}{\text{Coefficient of } x^2}$ [$1$ mark]
125open ↗
Find the zeroes of the polynomial $25a^2 - 10a + 1$ and verify the relationship between the zeroes and coefficients of the given polynomial.
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$25a^2 - 10a + 1 = (5a - 1) (5a - 1)$
Zeroes are $\frac{1}{5}$ and $\frac{1}{5}$
Sum of zeroes $= \frac{1}{5} + \frac{1}{5} = \frac{2}{5} = \frac{-(-10)}{25} = \frac{-\text{Coefficient of } a}{\text{Coefficient of } a^2}$
Product of zeroes $= \frac{1}{5} \times \frac{1}{5} = \frac{1}{25} = \frac{\text{Constant term}}{\text{Coefficient of } a^2}$
Zeroes are $\frac{1}{5}$ and $\frac{1}{5}$
Sum of zeroes $= \frac{1}{5} + \frac{1}{5} = \frac{2}{5} = \frac{-(-10)}{25} = \frac{-\text{Coefficient of } a}{\text{Coefficient of } a^2}$
Product of zeroes $= \frac{1}{5} \times \frac{1}{5} = \frac{1}{25} = \frac{\text{Constant term}}{\text{Coefficient of } a^2}$
126open ↗
Find the zeroes of the polynomial $p(x) = 6x^2 + 13x - 5$ and verify the relationship between its zeroes and the coefficients.
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$p(x) = 6x^2 + 13x - 5 = (3x - 1)(2x + 5)$
Zeroes of $p(x)$ are $\frac{1}{3}, -\frac{5}{2}$ [1 mark]
Sum of zeroes $= \frac{1}{3} - \frac{5}{2} = \frac{-13}{6} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$ [1 mark]
Product of zeroes $= \frac{1}{3} \times \frac{-5}{2} = \frac{-5}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ [1 mark]
Zeroes of $p(x)$ are $\frac{1}{3}, -\frac{5}{2}$ [1 mark]
Sum of zeroes $= \frac{1}{3} - \frac{5}{2} = \frac{-13}{6} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$ [1 mark]
Product of zeroes $= \frac{1}{3} \times \frac{-5}{2} = \frac{-5}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ [1 mark]
127open ↗
Find the zeroes of the polynomial $p(x) = 4x^2 - 4x - 3$ and verify the relationship between zeroes and its coefficients.
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$p(x) = 4x^2 - 4x - 3 = (2x + 1) (2x - 3)$
$\therefore$ zeroes of $p(x)$ are $-\frac{1}{2}$ and $\frac{3}{2}$ [$1$ mark]
Sum of zeroes $= -\frac{1}{2} + \frac{3}{2} = 1 = \frac{-(-4)}{4} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$ [$1$ mark]
Product of zeroes $= -\frac{1}{2} \times \frac{3}{2} = -\frac{3}{4} = \frac{\text{constant}}{\text{coefficient of } x^2}$ [$1$ mark]
$\therefore$ zeroes of $p(x)$ are $-\frac{1}{2}$ and $\frac{3}{2}$ [$1$ mark]
Sum of zeroes $= -\frac{1}{2} + \frac{3}{2} = 1 = \frac{-(-4)}{4} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$ [$1$ mark]
Product of zeroes $= -\frac{1}{2} \times \frac{3}{2} = -\frac{3}{4} = \frac{\text{constant}}{\text{coefficient of } x^2}$ [$1$ mark]
128open ↗
Find the zeroes of the polynomial $p(x) = 9x^2 - 6x - 35$ and verify the relationship between zeroes and its coefficients.
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$p(x) = 9x^2 - 6x - 35 = (3x - 7) (3x + 5)$
Zeroes of $p(x)$ are $\frac{7}{3}$ and $\frac{-5}{3}$
Sum of zeroes $= \frac{7}{3} - \frac{5}{3} = \frac{2}{3} = \frac{-(-6)}{9} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= \frac{7}{3} \times \frac{-5}{3} = \frac{-35}{9} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
Zeroes of $p(x)$ are $\frac{7}{3}$ and $\frac{-5}{3}$
Sum of zeroes $= \frac{7}{3} - \frac{5}{3} = \frac{2}{3} = \frac{-(-6)}{9} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= \frac{7}{3} \times \frac{-5}{3} = \frac{-35}{9} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
129open ↗
If $\alpha, \beta$ are zeroes of the polynomial $3x^2 - 8x + 4$, then form a quadratic polynomial in $x$ whose zeroes are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
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(a) $p(x) = 3x^2 - 8x + 4$
$\alpha + \beta = \frac{8}{3}, \alpha\beta = \frac{4}{3}$
$\therefore \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = 2$ and $\frac{1}{\alpha\beta} = \frac{3}{4}$
$\therefore$ required polynomial is $x^2 - 2x + \frac{3}{4}$ or $k(4x^2 - 8x + 3)$, where $k$ is a non-zero real number.
$\alpha + \beta = \frac{8}{3}, \alpha\beta = \frac{4}{3}$
$\therefore \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = 2$ and $\frac{1}{\alpha\beta} = \frac{3}{4}$
$\therefore$ required polynomial is $x^2 - 2x + \frac{3}{4}$ or $k(4x^2 - 8x + 3)$, where $k$ is a non-zero real number.
130open ↗
Find zeroes of the polynomial $6x^2 - 7x - 3$ and verify the relationship between zeroes and its coefficients.
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$p(x) = 6x^2 - 7x - 3 = (2x - 3)(3x + 1)$
Zeroes of $p(x)$ are $\frac{3}{2}$ and $-\frac{1}{3}$
Sum of zeroes $= \frac{3}{2} - \frac{1}{3} = \frac{7}{6} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= \frac{3}{2} \times \frac{-1}{3} = \frac{-3}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
Zeroes of $p(x)$ are $\frac{3}{2}$ and $-\frac{1}{3}$
Sum of zeroes $= \frac{3}{2} - \frac{1}{3} = \frac{7}{6} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= \frac{3}{2} \times \frac{-1}{3} = \frac{-3}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
131open ↗
If $\alpha, \beta$ are zeroes of the polynomial $8x^2 - 5x - 1$, then form a quadratic polynomial in $x$ whose zeroes are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$.
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$p(x) = 8x^2 - 5x - 1$
$\alpha + \beta = \frac{5}{8}, \alpha\beta = \frac{-1}{8}$ ($\frac{1}{2} + \frac{1}{2}$ marks)
$\therefore \text{sum of zeroes} = \frac{2}{\alpha} + \frac{2}{\beta} = -10$ ($\frac{1}{2}$ mark)
$\text{and product of zeroes} = \frac{2}{\alpha} \times \frac{2}{\beta} = -32$ ($\frac{1}{2}$ mark)
$\text{Required polynomial} = x^2 + 10x - 32$ (1 mark)
$\alpha + \beta = \frac{5}{8}, \alpha\beta = \frac{-1}{8}$ ($\frac{1}{2} + \frac{1}{2}$ marks)
$\therefore \text{sum of zeroes} = \frac{2}{\alpha} + \frac{2}{\beta} = -10$ ($\frac{1}{2}$ mark)
$\text{and product of zeroes} = \frac{2}{\alpha} \times \frac{2}{\beta} = -32$ ($\frac{1}{2}$ mark)
$\text{Required polynomial} = x^2 + 10x - 32$ (1 mark)
132open ↗
Find the zeroes of the polynomial $p(x) = 3x^2 + x - 10$ and verify the relationship between zeroes and its coefficients.
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$p(x) = 3x^2 + x - 10 = (x + 2)(3x - 5)$
$\text{Zeroes of } p(x) \text{ are } -2 \text{ and } \frac{5}{3}$ (1 mark)
$\text{Sum of zeroes} = -2 + \frac{5}{3} = \frac{-1}{3} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 mark)
$\text{Product of zeroes} = -2 \times \frac{5}{3} = \frac{-10}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1 mark)
$\text{Zeroes of } p(x) \text{ are } -2 \text{ and } \frac{5}{3}$ (1 mark)
$\text{Sum of zeroes} = -2 + \frac{5}{3} = \frac{-1}{3} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 mark)
$\text{Product of zeroes} = -2 \times \frac{5}{3} = \frac{-10}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1 mark)
133open ↗
$\alpha, \beta$ are zeroes of the polynomial $3x^2 - 8x + k$. Find the value of $k$, if $\alpha^2 + \beta^2 = \frac{40}{9}$.
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$p(x) = 3x^2 - 8x + k \Rightarrow \alpha + \beta = \frac{8}{3}, \alpha\beta = \frac{k}{3}$
$\alpha^2 + \beta^2 = \frac{40}{9} \Rightarrow (\frac{8}{3})^2 - \frac{2k}{3} = \frac{40}{9} \Rightarrow k = 4$
$\alpha^2 + \beta^2 = \frac{40}{9} \Rightarrow (\frac{8}{3})^2 - \frac{2k}{3} = \frac{40}{9} \Rightarrow k = 4$
134open ↗
Find the zeroes of the polynomial $2x^2 + 7x + 5$ and verify the relationship between its zeroes and co-efficients.
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$p(x) = 2x^2 + 7x + 5 = (x + 1)(2x + 5)$
Zeroes of $p(x)$ are $-1$ and $-\frac{5}{2}$
Sum of zeroes $= -1 - \frac{5}{2} = -\frac{7}{2} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= (-1)(-\frac{5}{2}) = \frac{5}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
Zeroes of $p(x)$ are $-1$ and $-\frac{5}{2}$
Sum of zeroes $= -1 - \frac{5}{2} = -\frac{7}{2} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$
Product of zeroes $= (-1)(-\frac{5}{2}) = \frac{5}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
135open ↗
Find the zeroes of the polynomial $p(x) = 3x^2 + 7x - 20$ and verify the relationship between its zeroes and the coefficients.
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$p(x) = 3x^2 + 7x - 20 = 3x^2 + 12x - 5x - 20 = 3x(x+4) - 5(x+4) = (3x-5)(x+4)$ (1 Mark)
Zeroes are $-4, \frac{5}{3}$ (1 Mark)
Sum of the zeroes = $-4 + \frac{5}{3} = \frac{-12+5}{3} = -\frac{7}{3} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 Mark)
Product of zeroes = $-4 \times \frac{5}{3} = -\frac{20}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1 Mark)
Zeroes are $-4, \frac{5}{3}$ (1 Mark)
Sum of the zeroes = $-4 + \frac{5}{3} = \frac{-12+5}{3} = -\frac{7}{3} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 Mark)
Product of zeroes = $-4 \times \frac{5}{3} = -\frac{20}{3} = \frac{\text{constant term}}{\text{coefficient of } x^2}$ (1 Mark)
136open ↗
Find the zeroes of the polynomial $p(x) = 4x^2 - 8x + 3$ and verify the relationship between its zeroes and co-efficients.
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$p(x) = 4x^2 - 8x + 3$
Zeroes of $p(x)$ are $\frac{3}{2}, \frac{1}{2}$ (1 Mark)
Sum of zeroes = $\frac{3}{2} + \frac{1}{2} = 2 = -(\frac{-8}{4}) = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 Mark)
Product of zeroes = $\frac{3}{2} \times \frac{1}{2} = \frac{3}{4} = \frac{\text{coefficient of } x^0}{\text{coefficient of } x^2}$ (1 Mark)
Zeroes of $p(x)$ are $\frac{3}{2}, \frac{1}{2}$ (1 Mark)
Sum of zeroes = $\frac{3}{2} + \frac{1}{2} = 2 = -(\frac{-8}{4}) = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (1 Mark)
Product of zeroes = $\frac{3}{2} \times \frac{1}{2} = \frac{3}{4} = \frac{\text{coefficient of } x^0}{\text{coefficient of } x^2}$ (1 Mark)
Making polynomials from zeroes
1 Mark Questions
137open ↗
Which of the following is a quadratic polynomial having zeroes $-\frac{2}{3}$ and $\frac{2}{3}$ ?
- (a)$4x^2-9$
- (b)$\frac{4}{9}(9x^2+4)$
- (c)$x^2 + \frac{9}{4}$
- (d)$5(9x^2-4)$
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(D) $5 (9x^2 – 4)$
138open ↗
Which of the following is a quadratic polynomial with zeroes $\frac{5}{3}$ and 0?
- (a)$3x (3x-5)$
- (b)$3x (x - 5)$
- (c)$x^2-\frac{5}{3}$
- (d)$\frac{5}{3} x^2$
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(A) $3x (3x - 5)$
139open ↗
The number of polynomials having zeroes $-3$ and $5$ is :
- (a)only one
- (b)Infinite
- (c)exactly two
- (d)at most two
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(b) Infinite
140open ↗
The number of polynomials having zeroes $-1$ and $2$ is :
- (a)exactly $2$
- (b)only $1$
- (c)at most $2$
- (d)infinite
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(d) infinite
141open ↗
The number of quadratic polynomials having zeroes $-5$ and $-3$ is
- (a)1
- (b)2
- (c)3
- (d)more than 3
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(D)more than 3
142open ↗
A quadratic polynomial, one of whose zeroes is $2 + \sqrt{5}$ and the sum of whose zeroes is $4$, is:
- (a)$x^2 + 4x-1$
- (b)$x^2 - 4x - 1$
- (c)$x^2 - 4x + 1$
- (d)$x^2 + 4x + 1$
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(B) $x^2 - 4x - 1$
143open ↗
A quadratic polynomial whose one zero is $3$ and the product of zeroes is $0$, is :
- (a)$x^2-3$
- (b)$x^2-9$
- (c)$x^2 + 3x$
- (d)$x^2 - 3x$
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(D) $x^2 - 3x$
144open ↗
The sum and product of zeroes of a quadratic polynomial $p(x)$ are $-\frac{1}{3}$ and $2$ respectively. The polynomial $p(x)$ is :
- (a)$3x^2 - x + 6$
- (b)$x^2 + \frac{1}{3}x - 2$
- (c)$3x^2 - x + 2$
- (d)$-3x^2 - x - 6$
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(D) $-3x^2 - x - 6$
145open ↗
If the zeroes of a polynomial $p(x)$ are $-3$ and $8$, then $p(x)$ equals
- (a)$x^2+5x-4$
- (b)$(x + 3)(-x+8)$
- (c)$a(x^2+5x-24)$
- (d)$x^2-24$
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(B) $(x + 3)(-x + 8)$ (1 Mark)
146open ↗
If sum and product of zeroes of a polynomial are $(-3)$ and $(-2)$ respectively, then a polynomial is
- (a)$x^2 - 3x - 2$
- (b)$-x^2 - 3x + 2$
- (c)$x^2 + 3x - 2$
- (d)$x^2 + 3x + 2$
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(B) $-x^2 - 3x + 2$
147open ↗
A quadratic polynomial having only zero $(-2)$ is
- (a)$(x - 2)^2$
- (b)$x^2 - 2$
- (c)$x^2 + 2x$
- (d)$(x + 2)^2$
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(D) $(x + 2)^2$
148open ↗
A quadratic polynomial having zeroes $0$ and $-2$, is
- (a)$x(x - 2)$
- (b)$4x(x + 2)$
- (c)$x^2 + 2$
- (d)$2x^2 + 2x$
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(B) $4x(x + 2)$
149open ↗
A quadratic polynomial, the sum and product of whose zeroes are $-1$ and $-2$ respectively, is
- (a)$x^2 - x - 2$
- (b)$2x^2 + x - 1$
- (c)$x^2 + x - 2$
- (d)$\frac{1}{2}x^2 + x - 4$
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(C) $x^2 + x - 2$
2 Marks Questions
150open ↗
Find a quadratic polynomial whose zeroes are $2$ and $-\frac{7}{5}$
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Sum of zeroes $= 2 + \left(-\frac{7}{5}\right) = \frac{3}{5}$ ($1/2$ mark)
Product of zeroes $= 2 \times \left(-\frac{7}{5}\right) = -\frac{14}{5}$ ($1/2$ mark)
$\therefore$ Required quadratic polynomial is $x^2 - \frac{3}{5}x - \frac{14}{5}$ or $5x^2 - 3x - 14$ ($1$ mark)
Product of zeroes $= 2 \times \left(-\frac{7}{5}\right) = -\frac{14}{5}$ ($1/2$ mark)
$\therefore$ Required quadratic polynomial is $x^2 - \frac{3}{5}x - \frac{14}{5}$ or $5x^2 - 3x - 14$ ($1$ mark)
151open ↗
Find a quadratic polynomial whose zeroes are $(5-2\sqrt{3})$ and $(5+2\sqrt{3})$.
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Let $\alpha$ and $\beta$ be the zeroes of the required polynomial.
$\alpha = 5 - 2\sqrt{3}, \beta = 5 + 2\sqrt{3}$
$\alpha + \beta = 10$ (I) (1/2 Mark)
$\alpha\beta = 13$ (II) (1/2 Mark)
$\therefore$ The quadratic polynomial is $x^2 - 10x + 13$ (III) (1 Mark)
$\alpha = 5 - 2\sqrt{3}, \beta = 5 + 2\sqrt{3}$
$\alpha + \beta = 10$ (I) (1/2 Mark)
$\alpha\beta = 13$ (II) (1/2 Mark)
$\therefore$ The quadratic polynomial is $x^2 - 10x + 13$ (III) (1 Mark)
152open ↗
Find the quadratic polynomial the sum of whose zeroes is $1$ and their product is $-12$. Hence find the zeroes of the polynomial.
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Sum of zeroes = $1$, Product of zeroes = $-12$ ($\frac{1}{2}$ Mark)
Required polynomial = $(x^2 - x - 12)$ (1 Mark)
$= (x - 4)(x + 3)$ ($\frac{1}{2}$ Mark)
Equating to zero, $x = 4, -3$
$\therefore$ Zeroes are $4$ and $-3$ ($\frac{1}{2}$ Mark)
Required polynomial = $(x^2 - x - 12)$ (1 Mark)
$= (x - 4)(x + 3)$ ($\frac{1}{2}$ Mark)
Equating to zero, $x = 4, -3$
$\therefore$ Zeroes are $4$ and $-3$ ($\frac{1}{2}$ Mark)
153open ↗
One zero of a quadratic polynomial is twice the other. If the sum of zeroes is (-6), find the polynomial.
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Let the zeroes be $\alpha, 2\alpha$
$\alpha + 2\alpha = -6$ gives $\alpha = -2$ (1/2 Mark)
Polynomial is $x^2 + 6x + 8$ (1 Mark)
$\alpha + 2\alpha = -6$ gives $\alpha = -2$ (1/2 Mark)
Polynomial is $x^2 + 6x + 8$ (1 Mark)
154open ↗
Form a quadratic polynomial whose zeroes are twice the zeroes of polynomial $p(x) = x^2 - 3x - 5$.
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Let $\alpha$ and $\beta$ be zeroes of $p(x)$
then $\alpha + \beta = 3$ and $\alpha\beta = -5$ (1)
Zeroes of required polynomial are $2\alpha$ and $2\beta$
$2\alpha + 2\beta = 6$ and $4\alpha\beta = -20$ (½)
$\therefore$ Required polynomial is $x^2 - 6x - 20$ or $k(x^2 - 6x - 20)$ (½)
then $\alpha + \beta = 3$ and $\alpha\beta = -5$ (1)
Zeroes of required polynomial are $2\alpha$ and $2\beta$
$2\alpha + 2\beta = 6$ and $4\alpha\beta = -20$ (½)
$\therefore$ Required polynomial is $x^2 - 6x - 20$ or $k(x^2 - 6x - 20)$ (½)
155open ↗
Form a quadratic polynomial whose sum and product of the zeroes are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively. Hence, find the zeroes of the polynomial.
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Required polynomial is $k(x^2 - \frac{1}{2}x - \frac{1}{2})$ or $18x^2 - 9x - 2$ (1 Mark)
Zeroes of the polynomial are $-\frac{1}{6}$ and $\frac{2}{3}$ (1 Mark)
Zeroes of the polynomial are $-\frac{1}{6}$ and $\frac{2}{3}$ (1 Mark)
3 Marks Questions
156open ↗
Find a quadratic polynomial whose sum of the zeroes is $8$ and difference of the zeroes is $2$.
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Let the zeroes be $\alpha$ and $\beta$
$\therefore \alpha + \beta = 8$ and $\alpha - \beta = 2$
Solving above two equations, we get $\alpha = 5$ and $\beta = 3$
So, the quadratic polynomial is $x^2 - 8x + 15$
$\therefore \alpha + \beta = 8$ and $\alpha - \beta = 2$
Solving above two equations, we get $\alpha = 5$ and $\beta = 3$
So, the quadratic polynomial is $x^2 - 8x + 15$
157open ↗
Find a quadratic polynomial whose sum and product of zeroes are $0$ and $- 9$, respectively. Also, find the zeroes of the polynomial so obtained.
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Polynomial is $x^2 - 0 (x) + (- 9) = x^2 - 9$
For zeroes :
$x^2 - 9 = (x + 3) (x - 3)$
Zeroes are $- 3, 3$
For zeroes :
$x^2 - 9 = (x + 3) (x - 3)$
Zeroes are $- 3, 3$
158open ↗
Find a quadratic polynomial, sum and product of whose zeroes are 5 and $- 6$, respectively. Also, find the zeroes of the polynomial so obtained.
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Required polynomial is $x^2 - 5x - 6$
For zeroes; $x^2 - 5x - 6 = (x - 6) (x + 1)$
Zeroes are $x = 6, - 1$
For zeroes; $x^2 - 5x - 6 = (x - 6) (x + 1)$
Zeroes are $x = 6, - 1$
159open ↗
Determine a quadratic polynomial, sum and product of whose zeroes are $-10$ and $24$, respectively. Also, determine the zeroes of the polynomial so obtained.
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Required polynomial is $x^2 + 10x + 24$
For zeroes: $x^2 + 10x + 24 = (x + 6) (x + 4)$
Zeroes are $-6, -4$
For zeroes: $x^2 + 10x + 24 = (x + 6) (x + 4)$
Zeroes are $-6, -4$
160open ↗
If $\alpha, \beta$ are zeroes of the polynomial $p(x) = 5x^2 - 7x - 3$, then form a quadratic polynomial whose zeroes are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$.
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$\alpha + \beta = -\frac{(-7)}{5} = \frac{7}{5}$, $\alpha\beta = \frac{-3}{5}$ (½ + ½ Mark)
Sum of zeroes of required polynomial = $\frac{2}{\alpha} + \frac{2}{\beta} = \frac{2(\alpha + \beta)}{\alpha\beta} = \frac{2(\frac{7}{5})}{-\frac{3}{5}} = -\frac{14}{3}$ (½ Mark)
Product of zeroes of the required polynomial = $\frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha\beta} = \frac{4}{-\frac{3}{5}} = -\frac{20}{3}$ (½ Mark)
Required polynomial is $k(x^2 - (\text{sum of zeroes})x + \text{product of zeroes})$
$x^2 - (-\frac{14}{3})x + (-\frac{20}{3}) = x^2 + \frac{14}{3}x - \frac{20}{3}$ or $k(3x^2 + 14x - 20)$ where $k$ is any non-zero real number. (1 Mark)
Sum of zeroes of required polynomial = $\frac{2}{\alpha} + \frac{2}{\beta} = \frac{2(\alpha + \beta)}{\alpha\beta} = \frac{2(\frac{7}{5})}{-\frac{3}{5}} = -\frac{14}{3}$ (½ Mark)
Product of zeroes of the required polynomial = $\frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha\beta} = \frac{4}{-\frac{3}{5}} = -\frac{20}{3}$ (½ Mark)
Required polynomial is $k(x^2 - (\text{sum of zeroes})x + \text{product of zeroes})$
$x^2 - (-\frac{14}{3})x + (-\frac{20}{3}) = x^2 + \frac{14}{3}x - \frac{20}{3}$ or $k(3x^2 + 14x - 20)$ where $k$ is any non-zero real number. (1 Mark)
161open ↗
Form a polynomial whose zeroes are $\alpha^2$ and $\beta^2$, where $\alpha$ and $\beta$ are zeroes of the polynomial $p(x) = x^2-3\sqrt{2}x+4$.
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$\alpha+\beta= 3\sqrt{2}$, $\alpha\beta = 4$ (1/2+1/2 Mark)
α² + β² = (α + β)² - 2αβ = 10 (1 Mark)
α² β² = 4² = 16 (1/2 Mark)
Required polynomial is $x^2 - 10x + 16$ or $k(x^2 - 10x + 16)$ where $k$ is a non zero real number. (1/2 Mark)
α² + β² = (α + β)² - 2αβ = 10 (1 Mark)
α² β² = 4² = 16 (1/2 Mark)
Required polynomial is $x^2 - 10x + 16$ or $k(x^2 - 10x + 16)$ where $k$ is a non zero real number. (1/2 Mark)
Real life application
4 Marks Questions
162open ↗
While playing in a garden, Samaira saw a honeycomb and asked her mother what is that. Her mother replied that it's a honeycomb made by honey bees to store honey. Also, she told her that the shape of the honeycomb formed is a mathematical structure. The mathematical representation of the honeycomb is shown in the graph.
Based on the above information, answer the following questions :
(i) How many zeroes are there for the polynomial represented by the graph given ?
(ii) Write the zeroes of the polynomial.
(iii) (a) If the zeroes of a polynomial $x^2 + (a + 1) x + b$ are $2$ and $-3$, then determine the values of $a$ and $b$.
OR
(iii) (b) If the square of difference of the zeroes of the polynomial $x^2 + px + 45$ is $144$, then find the value of $p$.
Based on the above information, answer the following questions :
(i) How many zeroes are there for the polynomial represented by the graph given ?
(ii) Write the zeroes of the polynomial.
(iii) (a) If the zeroes of a polynomial $x^2 + (a + 1) x + b$ are $2$ and $-3$, then determine the values of $a$ and $b$.
OR
(iii) (b) If the square of difference of the zeroes of the polynomial $x^2 + px + 45$ is $144$, then find the value of $p$.
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(i) Two
(ii) $7$ and $-7$
(iii) (a) $-(a + 1) = 2 + (-3) \Rightarrow a = 0$
$b = 2 \times (-3) \Rightarrow b = -6$
OR
(b) Let $\alpha$ and $\beta$ be the zeroes of given polynomial
Here, $\alpha + \beta = -p$ and $\alpha \beta = 45$
$(\alpha - \beta)^2 = 144$
$\Rightarrow (\alpha + \beta)^2 - 4\alpha\beta = 144$
$\Rightarrow (-p)^2 - 4 \times 45 = 144$
$\Rightarrow p = \pm 18$
(ii) $7$ and $-7$
(iii) (a) $-(a + 1) = 2 + (-3) \Rightarrow a = 0$
$b = 2 \times (-3) \Rightarrow b = -6$
OR
(b) Let $\alpha$ and $\beta$ be the zeroes of given polynomial
Here, $\alpha + \beta = -p$ and $\alpha \beta = 45$
$(\alpha - \beta)^2 = 144$
$\Rightarrow (\alpha + \beta)^2 - 4\alpha\beta = 144$
$\Rightarrow (-p)^2 - 4 \times 45 = 144$
$\Rightarrow p = \pm 18$
163open ↗
A ball is thrown in the air so that $t$ seconds after it is thrown, its height $h$ metre above its starting point is given by the polynomial $h = 25t - 5t^2$.
Observe the graph of the polynomial and answer the following questions :
(i) Write zeroes of the given polynomial.
(ii) Find the maximum height achieved by ball.
(iii) (a) After throwing upward, how much time did the ball take to reach to the height of $30$ m?
OR
(iii) (b) Find the two different values of $t$ when the height of the ball was $20$ m.
Observe the graph of the polynomial and answer the following questions :
(i) Write zeroes of the given polynomial.
(ii) Find the maximum height achieved by ball.
(iii) (a) After throwing upward, how much time did the ball take to reach to the height of $30$ m?
OR
(iii) (b) Find the two different values of $t$ when the height of the ball was $20$ m.
Show SolutionHide Solution↓
(i) Zeroes of the polynomial are $0$ and $5$
(ii) Maximum height achieved by ball
$= 25 \times \frac{5}{2} - 5 \times (\frac{5}{2})^2$
$= \frac{125}{4}$ or $31.25$ m
(iii) (a) $-5t^2 + 25t = 30$
$\Rightarrow t^2 - 5t + 6 = 0$
$\Rightarrow (t-2)(t-3) = 0$
$t \ne 3$, $t = 2$
OR
(iii) (b) $-5t^2 + 25t = 20$
$\Rightarrow t^2 - 5t + 4 = 0$
$\Rightarrow (t - 4)(t - 1) = 0$
$\Rightarrow t = 4, 1$
(ii) Maximum height achieved by ball
$= 25 \times \frac{5}{2} - 5 \times (\frac{5}{2})^2$
$= \frac{125}{4}$ or $31.25$ m
(iii) (a) $-5t^2 + 25t = 30$
$\Rightarrow t^2 - 5t + 6 = 0$
$\Rightarrow (t-2)(t-3) = 0$
$t \ne 3$, $t = 2$
OR
(iii) (b) $-5t^2 + 25t = 20$
$\Rightarrow t^2 - 5t + 4 = 0$
$\Rightarrow (t - 4)(t - 1) = 0$
$\Rightarrow t = 4, 1$
164open ↗
During a theatre drama, a backdrop of building arches was used. The
nshape of the curve shown below can be represented by the polynomial
n$p(x) = -x^2 + 2x + 8$, where $x$ is the length (in feet) on stage level.
n(i) Determine the height of the arch.
n(ii) (a) Find zeroes of the polynomial $p(x)$. Which points on the
ngraph represent the zeroes?
nOR
n(ii) (b) Find the span of the arch on the stage floor.
n(iii) Write the coordinates of the point of intersection of the above curve
nwith the $y$-axis.
nshape of the curve shown below can be represented by the polynomial
n$p(x) = -x^2 + 2x + 8$, where $x$ is the length (in feet) on stage level.
n(i) Determine the height of the arch.
n(ii) (a) Find zeroes of the polynomial $p(x)$. Which points on the
ngraph represent the zeroes?
nOR
n(ii) (b) Find the span of the arch on the stage floor.
n(iii) Write the coordinates of the point of intersection of the above curve
nwith the $y$-axis.
Show SolutionHide Solution↓
(i) $p (1) = - (1)^2 + 2 \times 1 + 8 = 9$ (1 Mark)
So, height of the arch is $9$ feet.
(ii) (a) $p(x) = -x^2 + 2x + 8$
$= - (x-4)(x + 2)$ (1 Mark)
$\therefore$ zeroes are $-2$ and $4$. (1/2 Mark)
Points $B$ and $A$ on the graph represent the zeroes. (1/2 Mark)
OR
(ii) (b) $p(x) = -x^2 + 2x + 8$
$= - (x - 4)(x + 2)$ (1/2 Mark)
$\therefore$ zeroes are $-2$ and $4$. (1/2 Mark)
Hence, Coordinates of point $A$ and $B$ are $(4, 0)$ and $(-2, 0)$ respectively.
Span of the arch on the stage floor, $AB = 4 + 2 = 6$ (1 Mark)
So, span of the arch on the stage floor is $6$ feet.
(iii) $p (0) = - (0)^2 + 2 \times 0 + 8 = 8$ (1/2 Mark)
$\therefore$ the given curve intersects $y$-axis at $(0, 8)$ (1/2 Mark)
So, height of the arch is $9$ feet.
(ii) (a) $p(x) = -x^2 + 2x + 8$
$= - (x-4)(x + 2)$ (1 Mark)
$\therefore$ zeroes are $-2$ and $4$. (1/2 Mark)
Points $B$ and $A$ on the graph represent the zeroes. (1/2 Mark)
OR
(ii) (b) $p(x) = -x^2 + 2x + 8$
$= - (x - 4)(x + 2)$ (1/2 Mark)
$\therefore$ zeroes are $-2$ and $4$. (1/2 Mark)
Hence, Coordinates of point $A$ and $B$ are $(4, 0)$ and $(-2, 0)$ respectively.
Span of the arch on the stage floor, $AB = 4 + 2 = 6$ (1 Mark)
So, span of the arch on the stage floor is $6$ feet.
(iii) $p (0) = - (0)^2 + 2 \times 0 + 8 = 8$ (1/2 Mark)
$\therefore$ the given curve intersects $y$-axis at $(0, 8)$ (1/2 Mark)
165open ↗
This section (Q. 36 to 38) has 3 case study based questions of 4 marks each.
An arch of a railway bridge, built on Chenab riverbed, is shown in the above diagram. It is a parabolic arch connecting two hills at P and Q. If the parabolic curve is represented by the polynomial $p(x) = -0.0025x^2 - 0.025x + 136$.
Observe the diagram and based on above information, answer the following questions :
(i) Write the co-ordinates of point A.
(ii) Find the span of the arch.
(iii) (a) Write the zeroes of the polynomial using diagram and verify the relationship between sum of zeroes and polynomials.
OR
(iii) (b) Find the values of $p(x)$ at $x = 100$ and $x = -100$. Are they same?
An arch of a railway bridge, built on Chenab riverbed, is shown in the above diagram. It is a parabolic arch connecting two hills at P and Q. If the parabolic curve is represented by the polynomial $p(x) = -0.0025x^2 - 0.025x + 136$.
Observe the diagram and based on above information, answer the following questions :
(i) Write the co-ordinates of point A.
(ii) Find the span of the arch.
(iii) (a) Write the zeroes of the polynomial using diagram and verify the relationship between sum of zeroes and polynomials.
OR
(iii) (b) Find the values of $p(x)$ at $x = 100$ and $x = -100$. Are they same?
Show SolutionHide Solution↓
(i) At $x = 0, p(x) = 136$
$\therefore$ Coordinates of point A = (0,136) (I) (1)
(ii) Span of the arch = $238.5 + 228.5 = 467$ units (I) (1)
(iii) (a) Zeroes of the polynomial are 228.5 and – 238.5 (I) (1)
Sum of zeroes = $-10 = \frac{-0.025}{-0.0025} = -\frac{\text{coefficient of x}}{\text{coefficient of x}^2}$ (II) (1)
OR
(iii) (b) $p(100) = 108.5$ (I) (1)
$p(-100) = 113.5$ (II) (1/2)
$\therefore p(100) \neq p(-100)$ (III) (1/2)
$\therefore$ Coordinates of point A = (0,136) (I) (1)
(ii) Span of the arch = $238.5 + 228.5 = 467$ units (I) (1)
(iii) (a) Zeroes of the polynomial are 228.5 and – 238.5 (I) (1)
Sum of zeroes = $-10 = \frac{-0.025}{-0.0025} = -\frac{\text{coefficient of x}}{\text{coefficient of x}^2}$ (II) (1)
OR
(iii) (b) $p(100) = 108.5$ (I) (1)
$p(-100) = 113.5$ (II) (1/2)
$\therefore p(100) \neq p(-100)$ (III) (1/2)
166open ↗
An arch of a railway bridge, built on Chenab riverbed, is shown in the above diagram. It is a parabolic arch connecting two hills at P and Q. If the parabolic curve is represented by the polynomial $p(x) = -0.0025x^2 - 0.025x + 136$.
Observe the diagram and based on above information, answer the following questions :
(i) Write the co-ordinates of point A.
(ii) Find the span of the arch.
(iii) (a) Write the zeroes of the polynomial using diagram and verify the relationship between sum of zeroes and polynomials.
OR
(iii) (b) Find the values of $p(x)$ at $x = 100$ and $x = -100$. Are they same?
Observe the diagram and based on above information, answer the following questions :
(i) Write the co-ordinates of point A.
(ii) Find the span of the arch.
(iii) (a) Write the zeroes of the polynomial using diagram and verify the relationship between sum of zeroes and polynomials.
OR
(iii) (b) Find the values of $p(x)$ at $x = 100$ and $x = -100$. Are they same?
Show SolutionHide Solution↓
(i) At $x = 0, p(x) = 136$
Coordinates of point A = $(0,136)$ (I Mark)
(ii) Span of the arch = $238.5 + 228.5 = 467$ units (I Mark)
(iii) (a) Zeroes of the polynomial are $228.5$ and $-238.5$ (I Mark)
Sum of zeroes = $-10 = \frac{-0.025}{-0.0025} = \frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (II Mark)
OR
(iii) (b) $p(100) = 108.5$ (I Mark)
$p(-100) = 113.5$ (II Mark)
$\therefore p(100) \neq p(-100)$ (III Mark)
Coordinates of point A = $(0,136)$ (I Mark)
(ii) Span of the arch = $238.5 + 228.5 = 467$ units (I Mark)
(iii) (a) Zeroes of the polynomial are $228.5$ and $-238.5$ (I Mark)
Sum of zeroes = $-10 = \frac{-0.025}{-0.0025} = \frac{\text{coefficient of } x}{\text{coefficient of } x^2}$ (II Mark)
OR
(iii) (b) $p(100) = 108.5$ (I Mark)
$p(-100) = 113.5$ (II Mark)
$\therefore p(100) \neq p(-100)$ (III Mark)
General
1 Mark Questions
167open ↗
Assertion (A): The polynomial $p(x) = x^2 + 3x + 3$ has two real zeroes.
Reason (R): A quadratic polynomial can have at most two real zeroes.
Reason (R): A quadratic polynomial can have at most two real zeroes.
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(d) Assertion (A) is false, but Reason (R) is true.