Previous-year board questions (2023–2026) with marking-scheme solutions, grouped by topic and marks.
Graphical Presentation of Zero 1 Mark Questions
9 1 Mark · March 2023 · Standard open ↗
The graph of $y = p(x)$ is given, for a polynomial $p(x)$. The number of zeroes of $p(x)$ from the graph is
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10 1 Mark · March 2023 · Standard open ↗
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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11 1 Mark · March 2024 · Standard open ↗
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 Solution Hide Solution ↓ (A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A)
12 1 Mark · March 2024 · Standard open ↗
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 Solution Hide Solution ↓ (d) Assertion (A) is false but Reason (R) is true.
13 1 Mark · March 2024 · Standard open ↗
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 :
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14 1 Mark · July 2025 · Standard open ↗
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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15 1 Mark · March 2025 · Standard open ↗
Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is:
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16 1 Mark · March 2026 · Standard open ↗
The graph of $y = f(x)$ is given. The number of zeroes of $f(x)$ is:
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17 1 Mark · March 2026 · Standard open ↗
The graph of $y = f(x)$ is given. The number of zeroes of $f(x)$ is :
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18 1 Mark · March 2026 · Standard open ↗
The graph of $y = f(x)$ is given. The number of distinct zeroes of $y = f(x)$ is :
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19 1 Mark · March 2026 · Standard open ↗
Observe the graph of polynomial $p(x)$. Number of zeroes of $p(x)$ is
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20 1 Mark · March 2026 · Standard open ↗
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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21 1 Mark · March 2025 · Basic open ↗
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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Find Zeroes 1 Mark Questions
22 1 Mark · 🔁 March 2023 & July 2023 · Standard open ↗
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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23 1 Mark · March 2023 · Standard open ↗
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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24 1 Mark · March 2024 · Standard open ↗
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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25 1 Mark · July 2025 · Standard open ↗
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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26 1 Mark · March 2025 · Standard open ↗
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 Solution Hide Solution ↓ (B) $2\sqrt{2}, \sqrt{2}$
27 1 Mark · March 2025 · Standard open ↗
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 Solution Hide Solution ↓ (D) $\frac{2}{3}, -\frac{1}{7}$
28 1 Mark · March 2025 · Standard open ↗
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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29 1 Mark · March 2026 · Standard open ↗
How many zeroes does $p(x) = (x-2)(x+3)$ have?
(a) Zero (b) One (c) Two (d) Three
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30 1 Mark · March 2026 · Standard open ↗
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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31 1 Mark · March 2026 · Standard open ↗
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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2 Marks Questions
32 2 Marks · March 2025 · Standard open ↗
Find the zeroes of the polynomial $p(x) = x^2 + \frac{4}{3}x - \frac{4}{3}$.
Show Solution Hide 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 $\frac{2}{3}, -2$.
33 2 Marks · March 2025 · Standard open ↗
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 Solution Hide 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 $\frac{6}{7}$
34 2 Marks · March 2026 · Basic open ↗
Find the zeroes of the polynomial $p(x) = 15x^2 – 19x + 6$.
Show Solution Hide 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
35 1 Mark · July 2023 · Standard open ↗
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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36 1 Mark · July 2023 · Standard open ↗
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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37 1 Mark · March 2023 · Standard open ↗
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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38 1 Mark · March 2023 · Standard open ↗
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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39 1 Mark · March 2023 · Standard open ↗
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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40 1 Mark · March 2023 · Standard open ↗
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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41 1 Mark · March 2023 · Standard open ↗
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 Solution Hide Solution ↓ (d) $a = \frac{1}{2}, c = 5$
42 1 Mark · March 2023 · Standard open ↗
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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43 1 Mark · March 2023 · Standard open ↗
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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44 1 Mark · March 2023 · Standard open ↗
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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45 1 Mark · July 2024 · Standard open ↗
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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46 1 Mark · March 2024 · Standard open ↗
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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47 1 Mark · March 2024 · Standard open ↗
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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48 1 Mark · March 2024 · Standard open ↗
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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49 1 Mark · March 2024 · Standard open ↗
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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50 1 Mark · March 2024 · Standard open ↗
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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51 1 Mark · March 2024 · Standard open ↗
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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52 1 Mark · March 2024 · Standard open ↗
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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53 1 Mark · 🔁 March 2024 & March 2025 · Standard open ↗
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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54 1 Mark · July 2025 · Standard open ↗
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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55 1 Mark · March 2025 · Standard open ↗
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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56 1 Mark · March 2025 · Standard open ↗
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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57 1 Mark · March 2025 · Standard open ↗
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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58 1 Mark · March 2025 · Standard open ↗
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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59 1 Mark · March 2025 · Standard open ↗
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) $-\frac{1}{2}$
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60 1 Mark · March 2025 · Standard open ↗
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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61 1 Mark · March 2026 · Standard open ↗
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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62 1 Mark · March 2026 · Standard open ↗
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 Solution Hide Solution ↓ (C) $3x^2 - 2x + 2$ (1 Mark)
63 1 Mark · March 2025 · Basic open ↗
$\alpha, \beta$ 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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64 1 Mark · March 2025 · Basic open ↗
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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65 1 Mark · March 2025 · Basic open ↗
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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66 1 Mark · March 2026 · Basic open ↗
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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67 1 Mark · March 2026 · Basic open ↗
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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68 1 Mark · March 2026 · Basic open ↗
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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2 Marks Questions
69 2 Marks · July 2023 · Standard open ↗
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 Solution Hide Solution ↓ $p + q = -\frac{1}{6}$, $pq = -\frac{2}{6} = -\frac{1}{3}$ $\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-\frac{1}{6}}{-\frac{1}{3}} = \frac{1}{2}$ $\frac{1}{p} + \frac{1}{q} - pq = \frac{1}{2} - (-\frac{1}{3}) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
70 2 Marks · March 2023 · Standard open ↗
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 Solution Hide Solution ↓ $p(x) = 6x^2 + 37x - (k – 2)$ Let the zeroes be $\alpha, \frac{1}{\alpha}$ Product of zeroes = $\alpha . \frac{1}{\alpha} = \frac{-(k-2)}{6}$ $1 = \frac{-(k-2)}{6}$ $6 = -k + 2 \Rightarrow k = -4$
71 2 Marks · March 2024 · Standard open ↗
If $\alpha, \beta$ are zeroes of the polynomial $p(x) = 5x^2 - 6x + 1$, then find the value of $\alpha + \beta + \alpha\beta$.
Show Solution Hide Solution ↓ $\alpha + \beta = \frac{6}{5}$ $\alpha\beta = \frac{1}{5}$ $\alpha + \beta + \alpha\beta = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}$
72 2 Marks · March 2024 · Standard open ↗
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 Solution Hide 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)
73 2 Marks · March 2025 · Standard open ↗
If the zeroes of the polynomial $x^2 + ax + b$ are in the ratio $3: 4$, then prove that $12a^2 = 49b$.
Show Solution Hide 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)
74 2 Marks · March 2025 · Standard open ↗
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 Solution Hide Solution ↓ $\alpha + \beta = 5$ $\alpha\beta = 3$ $\alpha^4\beta^3 + \alpha^3\beta^4 = (\alpha\beta)^3(\alpha + \beta)$ $= 27 \times 5 = 135$
75 2 Marks · March 2025 · Standard open ↗
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 Solution Hide Solution ↓ $$\begin{aligned}& \alpha + \beta = 2 \\ & \alpha\beta = -1 \\ & \frac{1}{2\alpha} + \frac{1}{2\beta} + 3\alpha\beta = \frac{\alpha + \beta}{2\alpha\beta} + 3\alpha\beta \\ & = \frac{2}{2(-1)} + 3(-1) = -4\end{aligned}$$
76 2 Marks · March 2026 · Standard open ↗
$\alpha$ and $\beta$ are the zeroes of the polynomial $5x^2 – 16x - 10$. Find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
Show Solution Hide 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)
77 2 Marks · March 2026 · Basic open ↗
Verify the relationship between the zeroes and the coefficients of the quadratic polynomial $25x^2 - 16$.
Show Solution Hide 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)
78 2 Marks · March 2026 · Basic open ↗
Verify the relationship between the zeroes and the coefficients of the quadratic polynomial $9x^2 - 25$.
Show Solution Hide 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)
79 2 Marks · March 2026 · Basic open ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $p(x) = 3x^2 – 5x$.
Show Solution Hide 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)
80 2 Marks · March 2026 · Basic open ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $2x^2 – 3x$.
Show Solution Hide 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)
81 2 Marks · March 2026 · Basic open ↗
Verify the relationship between the zeroes and the coefficients of the polynomial $5x^2 + 2x$.
Show Solution Hide 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)
82 2 Marks · March 2026 · Basic open ↗
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 Solution Hide 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
83 3 Marks · July 2024 · Standard open ↗
Find the zeroes of the polynomial $2t^2 - 9t - 45$ and verify the relationship between the zeroes and the coefficients of the polynomial.
Show Solution Hide 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 = $\frac{15}{2} \times (-3) = -\frac{45}{2} = \frac{\text{constant term}}{\text{coefficient of } t^2}$
84 3 Marks · March 2025 · Standard open ↗
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 Solution Hide 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 $2x^2 - 5x - 3$