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.

Year:Marks:Level:

Polynomial, Degree, Value, Find k

1 Mark Questions
11 Mark · March 2023 · Standardopen ↗
If one zero of the polynomial $x^2 + 3x + k$ is $2$, then the value of $k$.
  • (a)$-10$
  • (b)$10$
  • (c)$5$
  • (d)$-5$
Show SolutionHide Solution
(A) $-10$
21 Mark · March 2024 · Standardopen ↗
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)$16$
Show SolutionHide Solution
(C) $15$
31 Mark · March 2024 · Standardopen ↗
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)$5$
Show SolutionHide Solution
(B) $2$
41 Mark · March 2024 · Standardopen ↗
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)$-4$
Show SolutionHide Solution
(B) $4$
51 Mark · March 2024 · Standardopen ↗
Assertion (A): Degree of a zero polynomial is not defined.
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.
Show SolutionHide Solution
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
61 Mark · March 2025 · Basicopen ↗
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)$-\frac{22}{3}$
Show SolutionHide Solution
(d) $-\frac{22}{3}$
2 Marks Questions
72 Marks · March 2025 · Standardopen ↗
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'.
Show SolutionHide Solution
Sum of zeroes $$\begin{aligned}& = -\frac{2p + 3}{p+1} = -1 \\ & p = -2\end{aligned}$$
4 Marks Questions
84 Marks · March 2026 · Standardopen ↗
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.
Show SolutionHide Solution
(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)

Graphical Presentation of Zero

1 Mark Questions
91 Mark · March 2023 · Standardopen ↗
The graph of $y = p(x)$ is given, for a polynomial $p(x)$. The number of zeroes of $p(x)$ from the graph is
figure for this question
  • (a)3
  • (b)1
  • (c)2
  • (d)0
Show SolutionHide Solution
(B) 1
101 Mark · March 2023 · Standardopen ↗
The graph of $y = p(x)$ is given in the adjoining figure. Zeroes of the polynomial $p(x)$ are
figure for this question
  • (a)$-5, 7$
  • (b)$-\frac{5}{2}, -\frac{7}{2}$
  • (c)$-5, 0, 7$
  • (d)$-5, \frac{7}{2}, 7$
Show SolutionHide Solution
(C) $-5, 0, 7$
111 Mark · March 2024 · Standardopen ↗
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.
  • (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.
Show SolutionHide Solution
(A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A)
121 Mark · March 2024 · Standardopen ↗
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.
Show SolutionHide Solution
(d) Assertion (A) is false but Reason (R) is true.
131 Mark · March 2024 · Standardopen ↗
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)$0$
Show SolutionHide Solution
(B) $2$
141 Mark · July 2025 · Standardopen ↗
If the given figure shows the graph of polynomial $y = ax^2 + bx + c$, then :
figure for this question
  • (a)$a < 0$
  • (b)$b^2 < 4ac$
  • (c)$c > 0$
  • (d)a and b are of same sign
Show SolutionHide Solution
(A) $a < 0$
151 Mark · March 2025 · Standardopen ↗
Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is:
figure for this question
  • (a)$3$
  • (b)$5$
  • (c)$2$
  • (d)$4$
Show SolutionHide Solution
(C) $2$
161 Mark · March 2026 · Standardopen ↗
The graph of $y = f(x)$ is given.
The number of zeroes of $f(x)$ is:
figure for this question
  • (a)$0$
  • (b)$1$
  • (c)$2$
  • (d)$4$
Show SolutionHide Solution
(a) $0$
171 Mark · March 2026 · Standardopen ↗
The graph of $y = f(x)$ is given.
The number of zeroes of $f(x)$ is :
figure for this question
  • (a)$0$
  • (b)$1$
  • (c)$2$
  • (d)$4$
Show SolutionHide Solution
(a) $0$ (1 Mark)
181 Mark · March 2026 · Standardopen ↗
The graph of $y = f(x)$ is given. The number of distinct zeroes of $y = f(x)$ is :
figure for this question
  • (a)$0$
  • (b)$1$
  • (c)$2$
  • (d)$3$
Show SolutionHide Solution
(C) $2$ (1 Mark)
191 Mark · March 2026 · Standardopen ↗
Observe the graph of polynomial $p(x)$. Number of zeroes of $p(x)$ is
figure for this question
  • (a)$5$
  • (b)$4$
  • (c)$6$
  • (d)$3$
Show SolutionHide Solution
(D) $3$
201 Mark · March 2026 · Standardopen ↗
Observe the graph of polynomial $p(x)$. The zeroes of the polynomial are
figure for this question
  • (a)$-2, 0, 2.5$
  • (b)$-2, 2.5$
  • (c)$0,4$
  • (d)$-2,0$
Show SolutionHide Solution
(B) $-2, 2.5$
211 Mark · March 2025 · Basicopen ↗
The graph of the polynomial $ax^2 + bx + c$ is a downward parabola if
  • (a)$a > 0$
  • (b)$a < 0$
  • (c)$a = 0$
  • (d)$a > 1$
Show SolutionHide Solution
(B) $a < 0$

Find Zeroes

1 Mark Questions
221 Mark · 🔁 March 2023 & July 2023 · Standardopen ↗
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)$\frac{1}{3}$, $4$
Show SolutionHide Solution
(c) $\frac{1}{3}$, $-4$
231 Mark · March 2023 · Standardopen ↗
The zeroes of the polynomial $p(x) = x^2 + 4x + 3$ are given by:
  • (a)$1,3$
  • (b)$-1,3$
  • (c)$1,-3$
  • (d)$-1,-3$
Show SolutionHide Solution
(d) $-1, -3$
241 Mark · March 2024 · Standardopen ↗
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)$3,\frac{2}{2}$
Show SolutionHide Solution
(A) $3, -\frac{3}{2}$
251 Mark · July 2025 · Standardopen ↗
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
Show SolutionHide Solution
(B) both negative
261 Mark · March 2025 · Standardopen ↗
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)$\sqrt{2},2$
Show SolutionHide Solution
(B) $2\sqrt{2}, \sqrt{2}$
271 Mark · March 2025 · Standardopen ↗
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)$\frac{2}{3}, -\frac{1}{7}$
Show SolutionHide Solution
(D) $\frac{2}{3}, -\frac{1}{7}$
281 Mark · March 2025 · Standardopen ↗
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)$-3$
Show SolutionHide Solution
(D) $-3$
291 Mark · March 2026 · Standardopen ↗
How many zeroes does $p(x) = (x-2)(x+3)$ have?
  • (a)Zero
  • (b)One
  • (c)Two
  • (d)Three
Show SolutionHide Solution
(C) Two
301 Mark · March 2026 · Standardopen ↗
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)$(\frac{1}{3}, 0)$
Show SolutionHide Solution
(D) $(\frac{1}{3}, 0)$
311 Mark · March 2026 · Standardopen ↗
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)$(x - 1), (x + 2)$
Show SolutionHide Solution
(A) $(x + 2), (x - 5)$
2 Marks Questions
322 Marks · March 2025 · Standardopen ↗
Find the zeroes of the polynomial $p(x) = x^2 + \frac{4}{3}x - \frac{4}{3}$.
Show SolutionHide Solution
$\frac{1}{3}(3x^2 + 4x - 4) = \frac{1}{3}(3x^2 + 6x - 2x - 4) = \frac{1}{3}(3x - 2)(x + 2)$.
Zeroes are $\frac{2}{3}, -2$.
332 Marks · March 2025 · Standardopen ↗
If $p$ and $q$ are zeroes of the polynomial $p(y) = 21y^2 - y - 2$, then find the value of $(1-p) . (1 - q)$.
Show SolutionHide Solution
$p+q=\frac{1}{21}$
$p.q=\frac{-2}{21}$
$(1-p)(1 - q) = 1 - (p + q) + pq$
$= 1-\frac{1}{21} - \frac{2}{21}$
$= \frac{18}{21}$ or $\frac{6}{7}$
342 Marks · March 2026 · Basicopen ↗
Find the zeroes of the polynomial $p(x) = 15x^2 – 19x + 6$.
Show SolutionHide Solution
For zeroes $15x^2 – 19x + 6= 0$ (1 Mark)
Zeroes are $\frac{3}{5}$ and $\frac{2}{3}$ (1 Mark)

Relationship of Zeros and Coefficients

1 Mark Questions
351 Mark · July 2023 · Standardopen ↗
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)$37$
Show SolutionHide Solution
(b) $\frac{37}{4}$
361 Mark · July 2023 · Standardopen ↗
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)$-\frac{3}{2}$
Show SolutionHide Solution
(a) $-\frac{2}{3}$
371 Mark · March 2023 · Standardopen ↗
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)$-\frac{1}{2}$
Show SolutionHide Solution
(A) 1
381 Mark · March 2023 · Standardopen ↗
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)$\frac{3}{4}$
Show SolutionHide Solution
(B) $\frac{5}{4}$
391 Mark · March 2023 · Standardopen ↗
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)$b^2 + 2a$
Show SolutionHide Solution
(b) $a^2 + 2b$
401 Mark · March 2023 · Standardopen ↗
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)$-2$
Show SolutionHide Solution
(b) $2$
411 Mark · March 2023 · Standardopen ↗
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)$a = \frac{1}{2}, c = 5$
Show SolutionHide Solution
(d) $a = \frac{1}{2}, c = 5$
421 Mark · March 2023 · Standardopen ↗
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)$1$
Show SolutionHide Solution
(c) $0$
431 Mark · March 2023 · Standardopen ↗
If $\alpha$, $\beta$ are zeroes of the polynomial $x^2-1$, then value of $(\alpha + \beta)$ is :
  • (a)$2$
  • (b)$1$
  • (c)$-1$
  • (d)$0$
Show SolutionHide Solution
(d) $0$
441 Mark · March 2023 · Standardopen ↗
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)$-\frac{3}{7}$
Show SolutionHide Solution
(d) $-\frac{3}{7}$
451 Mark · July 2024 · Standardopen ↗
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)$-1$
Show SolutionHide Solution
(D) $-1$
461 Mark · March 2024 · Standardopen ↗
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)$\frac{1}{2}$
Show SolutionHide Solution
(b) $2$
471 Mark · March 2024 · Standardopen ↗
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)$10$
Show SolutionHide Solution
(a) $-\frac{5}{2}$
481 Mark · March 2024 · Standardopen ↗
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)$\frac{5}{24}$
Show SolutionHide Solution
(b) $-\frac{5}{4}$
491 Mark · March 2024 · Standardopen ↗
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)$\frac{5}{-7}$
Show SolutionHide Solution
(C) $\frac{3}{7}$
501 Mark · March 2024 · Standardopen ↗
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)$\frac{71}{4}$
Show SolutionHide Solution
(B) $\frac{61}{4}$
511 Mark · March 2024 · Standardopen ↗
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
Show SolutionHide Solution
(B) 10
521 Mark · March 2024 · Standardopen ↗
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)$7:2$
Show SolutionHide Solution
(B) $2:7$
531 Mark · 🔁 March 2024 & March 2025 · Standardopen ↗
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)$\frac{2}{3}$
Show SolutionHide Solution
(D)$\frac{2}{3}$
541 Mark · July 2025 · Standardopen ↗
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)$-\frac{1}{2}$
Show SolutionHide Solution
(C)-$\frac{5}{2}$
551 Mark · March 2025 · Standardopen ↗
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)$4$
Show SolutionHide Solution
(D) $4$
561 Mark · March 2025 · Standardopen ↗
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)$-\frac{5}{7}$
Show SolutionHide Solution
(C) $\frac{5}{7}$
571 Mark · March 2025 · Standardopen ↗
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)$2$
Show SolutionHide Solution
(D) $2$
581 Mark · March 2025 · Standardopen ↗
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)$-a+b$
Show SolutionHide Solution
(C) $a - b$
591 Mark · March 2025 · Standardopen ↗
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
601 Mark · March 2025 · Standardopen ↗
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)$\pm 12$
Show SolutionHide Solution
(B) $\pm 15$
611 Mark · March 2026 · Standardopen ↗
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)$-\frac{1}{3}$
Show SolutionHide Solution
(B) $\frac{2}{3}$
621 Mark · March 2026 · Standardopen ↗
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)$x^2 - 3x + 2$
Show SolutionHide Solution
(C) $3x^2 - 2x + 2$ (1 Mark)
631 Mark · March 2025 · Basicopen ↗
$\alpha, \beta$ 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)$-5$
Show SolutionHide Solution
(d) $-5$
641 Mark · March 2025 · Basicopen ↗
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)$-\frac{42}{5}$
Show SolutionHide Solution
(b) $\frac{42}{5}$
651 Mark · March 2025 · Basicopen ↗
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)$4$
Show SolutionHide Solution
(B) $-\frac{1}{4}$
661 Mark · March 2026 · Basicopen ↗
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)$6$
Show SolutionHide Solution
$6$
671 Mark · March 2026 · Basicopen ↗
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)$\frac{7}{6}$
Show SolutionHide Solution
$\frac{7}{6}$
681 Mark · March 2026 · Basicopen ↗
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)$\frac{30}{11}$
Show SolutionHide Solution
(B) $\frac{11}{30}$
2 Marks Questions
692 Marks · July 2023 · Standardopen ↗
If $p$ and $q$ are the zeroes of the quadratic polynomial $f(x) = 6x^2 + x - 2$, then find the value of $\frac{1}{p} + \frac{1}{q} - pq$.
Show SolutionHide Solution
$p + q = -\frac{1}{6}$, $pq = -\frac{2}{6} = -\frac{1}{3}$
$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-\frac{1}{6}}{-\frac{1}{3}} = \frac{1}{2}$
$\frac{1}{p} + \frac{1}{q} - pq = \frac{1}{2} - (-\frac{1}{3}) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
702 Marks · March 2023 · Standardopen ↗
If one zero of the polynomial $p(x) = 6x^2 + 37x – (k – 2)$ is reciprocal of the other, then find the value of $k$.
Show SolutionHide Solution
$p(x) = 6x^2 + 37x - (k – 2)$
Let the zeroes be $\alpha, \frac{1}{\alpha}$
Product of zeroes = $\alpha . \frac{1}{\alpha} = \frac{-(k-2)}{6}$
$1 = \frac{-(k-2)}{6}$
$6 = -k + 2 \Rightarrow k = -4$
712 Marks · March 2024 · Standardopen ↗
If $\alpha, \beta$ are zeroes of the polynomial $p(x) = 5x^2 - 6x + 1$, then find the value of $\alpha + \beta + \alpha\beta$.
Show SolutionHide Solution
$\alpha + \beta = \frac{6}{5}$
$\alpha\beta = \frac{1}{5}$
$\alpha + \beta + \alpha\beta = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}$
722 Marks · March 2024 · Standardopen ↗
If $\alpha$ and $\beta$ are zeroes of the quadratic polynomial $p(x) = x^2 - 5x + 4$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta$.
Show SolutionHide Solution
$\alpha + \beta = 5$ (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)
732 Marks · March 2025 · Standardopen ↗
If the zeroes of the polynomial $x^2 + ax + b$ are in the ratio $3: 4$, then prove that $12a^2 = 49b$.
Show SolutionHide Solution
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)
742 Marks · March 2025 · Standardopen ↗
If '$\alpha$' and '$\beta$' are the zeroes of the polynomial $p(y) = y^2 - 5y + 3$, then find the value of $\alpha^4\beta^3 + \alpha^3\beta^4$.
Show SolutionHide Solution
$\alpha + \beta = 5$
$\alpha\beta = 3$
$\alpha^4\beta^3 + \alpha^3\beta^4 = (\alpha\beta)^3(\alpha + \beta)$
$= 27 \times 5 = 135$
752 Marks · March 2025 · Standardopen ↗
If $\alpha$ and $\beta$ are zeroes of the polynomial $p(x) = x^2 - 2x - 1$, then find the value of $\frac{1}{2\alpha} + \frac{1}{2\beta} + 3\alpha\beta$.
Show SolutionHide Solution
$$\begin{aligned}& \alpha + \beta = 2 \\ & \alpha\beta = -1 \\ & \frac{1}{2\alpha} + \frac{1}{2\beta} + 3\alpha\beta = \frac{\alpha + \beta}{2\alpha\beta} + 3\alpha\beta \\ & = \frac{2}{2(-1)} + 3(-1) = -4\end{aligned}$$
762 Marks · March 2026 · Standardopen ↗
$\alpha$ and $\beta$ are the zeroes of the polynomial $5x^2 – 16x - 10$. Find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
Show SolutionHide Solution
$\alpha + \beta = \frac{16}{5}$, $\alpha \beta = -2$ (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)
772 Marks · March 2026 · Basicopen ↗
Verify the relationship between the zeroes and the coefficients of the quadratic polynomial $25x^2 - 16$.
Show SolutionHide Solution
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)
782 Marks · March 2026 · Basicopen ↗
Verify the relationship between the zeroes and the coefficients of the quadratic polynomial $9x^2 - 25$.
Show SolutionHide Solution
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)
792 Marks · March 2026 · Basicopen ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $p(x) = 3x^2 – 5x$.
Show SolutionHide Solution
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)
802 Marks · March 2026 · Basicopen ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $2x^2 – 3x$.
Show SolutionHide Solution
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)
812 Marks · March 2026 · Basicopen ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $5x^2 + 2x$.
Show SolutionHide Solution
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)
822 Marks · March 2026 · Basicopen ↗
If $\alpha$, $\beta$ are the zeroes of polynomial $p(x) = 6x^2-5x - 3$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
Show SolutionHide Solution
(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)
3 Marks Questions
833 Marks · July 2024 · Standardopen ↗
Find the zeroes of the polynomial $2t^2 - 9t - 45$ and verify the relationship between the zeroes and the coefficients of the polynomial.
Show SolutionHide Solution
$2t^2 - 9t - 45 = 2t^2 - 15t + 6t - 45$
$= (2t - 15) (t + 3)$
$\therefore$ 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 = $\frac{15}{2} \times (-3) = -\frac{45}{2} = \frac{\text{constant term}}{\text{coefficient of } t^2}$
843 Marks · March 2025 · Standardopen ↗
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 $p(x)$.
Show SolutionHide Solution
$p(x) = 6x^2 - 5x - 1 = (x - 1)(6x + 1)$
$\therefore$ 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 $2x^2 - 5x - 3$

Real life application

4 Marks Questions
854 Marks · July 2023 · Standardopen ↗
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 $p$.
figure for this question
Show SolutionHide Solution
(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 $\alpha \beta = 45$
$(\alpha - \beta)^2 = 144$
$\Rightarrow (\alpha + \beta)^2 - 4\alpha\beta = 144$
$\Rightarrow (-p)^2 - 4 \times 45 = 144$
$\Rightarrow p = \pm 18$