104 previous-year board questions (2023–2025) with marking-scheme solutions, grouped by topic and marks.
Relationship of Zeros and Coefficients
1 Mark Questions
321 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)$37$
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331 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)$-\frac{3}{2}$
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341 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)$-\frac{1}{2}$
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351 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)$\frac{3}{4}$
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361 Mark · March 2023 · Standardopen ↗
If one zero of the polynomial $6x^2 + 37x - (k-2)$ is reciprocal of the other, then what is the value of k?
- (a)$-4$
- (b)$-6$
- (c)$6$
- (d)$4$
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371 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)$b^2 + 2a$
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381 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)$-2$
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391 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)$a = \frac{1}{2}, c = 5$
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(d) $a = \frac{1}{2}, c = 5$
401 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)$1$
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411 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)$0$
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421 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)$-\frac{3}{7}$
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431 Mark · March 2023 · Standardopen ↗
If $\alpha$, $\beta$ are the zeroes of the polynomial $p(x) = 4x^2 - 3x - 7$, then $\left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$ is equal to :
- (a)$\frac{7}{3}$
- (b)$-\frac{7}{3}$
- (c)$\frac{3}{7}$
- (d)$-\frac{3}{7}$
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441 Mark · March 2023 · Standardopen ↗
If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 - 1$, then the value of $(\alpha + \beta)$ is
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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)$-1$
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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)$\frac{1}{2}$
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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)$10$
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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)$\frac{5}{24}$
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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)$\frac{3}{-7}$
- (c)$\frac{3}{5}$
- (d)$\frac{5}{-7}$
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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)$1$
- (c)$\frac{61}{4}$
- (d)$\frac{71}{4}$
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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
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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)$21:5$
- (c)$2:7$
- (d)$7:2$
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531 Mark · March 2024 · Standardopen ↗
If $\alpha$ and $\beta$ are the 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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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)$-\frac{1}{2}$
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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)$4$
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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)$-\frac{5}{7}$
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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)$2$
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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)$-a+b$
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591 Mark · 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)$\frac{2}{3}$
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601 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)$-\frac{1}{2}$
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611 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)$\pm 12$
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2 Marks Questions
622 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 $\frac{1}{p} + \frac{1}{q} - pq$.
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$p + q = -\frac{1}{6}$, $pq = -\frac{2}{6} = -\frac{1}{3}$
$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-\frac{1}{6}}{-\frac{1}{3}} = \frac{1}{2}$
$\frac{1}{p} + \frac{1}{q} - pq = \frac{1}{2} - (-\frac{1}{3}) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
632 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 $k$.
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Sol. $p(x) = 6x^2 + 37x - (k – 2)$
Let the zeroes be $\alpha, \frac{1}{\alpha}$
Product of zeroes = $\alpha . \frac{1}{\alpha} = \frac{-(k-2)}{6}$
$1 = \frac{-(k-2)}{6}$
$6 = -k + 2 \Rightarrow k = -4$
642 Marks · March 2024 · Standardopen ↗
If $\alpha, \beta$ are zeroes of the polynomial $p(x) = 5x^2 - 6x + 1$, then find the value of $\alpha + \beta + \alpha\beta$.
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$\alpha + \beta = \frac{6}{5}$
$\alpha\beta = \frac{1}{5}$
$\alpha + \beta + \alpha\beta = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}$
652 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 $\frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta$.
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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)
662 Marks · March 2025 · Standardopen ↗
If the zeroes of the polynomial $x^2 + ax + b$ are in the ratio $3: 4$, then prove that $12a^2 = 49b$.
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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)
672 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 $\alpha^4\beta^3 + \alpha^3\beta^4$.
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Sol. $\alpha + \beta = 5$
$\alpha\beta = 3$
$\alpha^4\beta^3 + \alpha^3\beta^4 = (\alpha\beta)^3(\alpha + \beta)$
$= 27 \times 5 = 135$
682 Marks · March 2025 · Standardopen ↗
If $\alpha$ and $\beta$ are zeroes of the polynomial $p(x) = x^2 - 2x - 1$, then find the value of $\frac{1}{2\alpha} + \frac{1}{2\beta} + 3\alpha\beta$.
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$$\begin{aligned}& \alpha + \beta = 2 \\ & alpha\beta = -1 \\ & frac{1}{2\alpha} + \frac{1}{2\beta} + 3\alpha\beta = \frac{\alpha + \beta}{2\alpha\beta} + 3\alpha\beta \\ & = \frac{2}{2(-1)} + 3(-1) = -4\end{aligned}$$
3 Marks Questions
693 Marks · March 2023 · Standardopen ↗
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. $x^2 - 29x + 100 = 0$
703 Marks · March 2023 · Standardopen ↗
OR
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 $\alpha, 3\alpha$
$\alpha + 3\alpha = -12 \Rightarrow \alpha =-3$
$\alpha.3\alpha=-k \Rightarrow 3\alpha^2 = -k$
$\Rightarrow k = -27$
713 Marks · July 2024 · Standardopen ↗
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, $\alpha + \beta = \frac{\alpha\beta}{3}$
$k+5 = \frac{5k+1}{3}$
$\Rightarrow k = 7$
723 Marks · July 2024 · Standardopen ↗
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 $3x^2 - 13x - 10$
$\therefore \alpha + \beta = \frac{13}{3}$, $\alpha\beta = \frac{-10}{3}$
$(3\alpha + 1). (3\beta + 1) = 9\alpha\beta + 3(\alpha + \beta) + 1$
$= -30 + 13 + 1$
$= -16$
733 Marks · July 2024 · Standardopen ↗
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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Sol. $f(x) = 6x^2 + 11x - 10$
$\alpha + \beta = -\frac{11}{6}$ and $\alpha\beta = -\frac{10}{6}$
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$
$= \frac{(-\frac{11}{6})^2 - 2(-\frac{10}{6})}{-\frac{10}{6}}$
$= -\frac{241}{60}$
743 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.
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Sol. $2t^2 - 9t - 45 = 2t^2 - 15t + 6t - 45$
$= (2t - 15) (t + 3)$
$\therefore$ 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 = $\frac{15}{2} \times (-3) = -\frac{45}{2} = \frac{\text{constant term}}{\text{coefficient of } t^2}$
753 Marks · March 2024 · Standardopen ↗
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 = $-\sqrt{15} \times \sqrt{15} = -15 = \frac{-15}{1} = \frac{\text{constant term}}{\text{coefficient of } x^2}$
763 Marks · March 2024 · Standardopen ↗
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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$$\begin{aligned}& P(x) = 4x^2 + 4x - 3 \\ & = (2x + 3) (2x - 1) \\ & \therefore \text{Zeroes of the polynomial are } -\frac{3}{2}, \frac{1}{2} \\ & \text{Sum of Zeroes } = -\frac{3}{2} + \frac{1}{2} = \frac{-3+1}{2} = \frac{-4}{4} = -1 = -\frac{(\text{coefficient of } x)}{(\text{coefficient of } x^2)} \\ & \text{Product of Zeroes } = -\frac{3}{2} \times \frac{1}{2} = -\frac{3}{4} = \frac{\text{constant term}}{(\text{coefficient of } x^2)}\end{aligned}$$
773 Marks · March 2024 · Standardopen ↗
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 $$\begin{aligned}& \alpha\beta = -2 \\ & \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} \\ & = \frac{(-1)^2-2(-2)}{-2} = \frac{1+4}{-2} = -\frac{5}{2}\end{aligned}$$
783 Marks · March 2024 · Standardopen ↗
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 $$\begin{aligned}& \alpha\beta = -\frac{10}{6} \\ & \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} \\ & = \frac{(-\frac{11}{6})^2-2\times(-\frac{10}{6})}{-\frac{10}{6}} \\ & = \frac{\frac{121}{36}+\frac{20}{6}}{-\frac{10}{6}} = \frac{\frac{121+120}{36}}{-\frac{10}{6}} = \frac{241}{36} \times -\frac{6}{10} = -\frac{241}{60}\end{aligned}$$
793 Marks · March 2024 · Standardopen ↗
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
sqrt{3} +
sqrt{3}) = -4
sqrt{3} = -
frac{coefficient of t}{coefficient of t^2}
Product of the zeroes = $(-5\sqrt{3}) \times \sqrt{3}) = -15 = \frac{\text{constant term}}{\text{coefficient of t}^2}$
803 Marks · July 2025 · Standardopen ↗
$\alpha, \beta$ 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 $\alpha \beta = k$
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$
$\Rightarrow (8)^2 = (4)^2 - 4k$
$\Rightarrow k = -12$
813 Marks · July 2025 · Standardopen ↗
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.
823 Marks · July 2025 · Standardopen ↗
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 $x^2 - 10x + 21$
833 Marks · March 2025 · Standardopen ↗
$\alpha$ 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 $x^2 + 2qx + 4p$
843 Marks · March 2025 · Standardopen ↗
$\alpha$ 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 $x^2 - (3a + 2)x + (3a - 9b + 1)$
853 Marks · March 2025 · Standardopen ↗
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 $ax^2 - (1 - 6a)x + (c - 3 + 9a)$.
863 Marks · March 2025 · Standardopen ↗
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 $7x^2 + 36x - 36$
873 Marks · March 2025 · Standardopen ↗
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 $2x^2 - x - 6$
883 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 $p(x)$.
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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 $2x^2 - 5x - 3$
893 Marks · March 2025 · Standardopen ↗
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).
903 Marks · March 2025 · Standardopen ↗
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 $8x^2 + 30x + 25$.
913 Marks · March 2025 · Standardopen ↗
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 $(x^2 - 3x - 4)$