Polynomials — Class 10 Maths PYQs

104 previous-year board questions (2023–2025) 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
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)$5$
  • (c)$10$
  • (d)$-5$
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(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)$15$
  • (c)$14$
  • (d)$16$
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(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)$4$
  • (c)$2$
  • (d)$5$
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Sol.
(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)$2$
  • (c)$4$
  • (d)$-4$
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(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.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
61 Mark · March 2024 · Standardopen ↗
Questions number $19$ and $20$ are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.
(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.
Assertion (A): Degree of a zero polynomial is not defined.
Reason (R): Degree of a non-zero constant polynomial is $0$.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
71 Mark · March 2025 · Standardopen ↗
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) $9$
81 Mark · March 2025 · Standardopen ↗
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.
91 Mark · March 2025 · Standardopen ↗
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.
2 Marks Questions
102 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'.
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Sum of zeroes $$\begin{aligned}& = -\frac{2p + 3}{p+1} = -1 \\ & p = -2\end{aligned}$$

Graphical Presentation of Zero

1 Mark Questions
111 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
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(B) 1
121 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)$-5, 0, 7$
  • (c)$-\frac{5}{2}, -\frac{7}{2}$
  • (d)$-5, \frac{7}{2}, 7$
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(C) $-5, 0, 7$
131 Mark · March 2024 · Standardopen ↗
Directions : In Question $19$ and $20$, Assertion (A) and Reason (R) are given. Select the correct option from the following :
(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.
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.
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(A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A)
141 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.
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Sol. (d) Assertion (A) is false but Reason (R) is true.
151 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)$0$
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(B) $2$
161 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
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(A) $a<0$
171 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)$c > 0$
  • (c)$b^2 < 4ac$
  • (d)a and b are of same sign
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(A) $a < 0$
181 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)$4$
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(C) $2$

Find Zeroes

1 Mark Questions
191 Mark · 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)$\frac{1}{3}, \frac{1}{4}$
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(c) $\frac{1}{3}, -4$
201 Mark · July 2023 · Standardopen ↗
The zeroes of the polynomial $3x^2 + 11x-4$ are:
  • (a)$\frac{1}{2}$, $-4$
  • (b)$\frac{1}{3}$, $-4$
  • (c)$\frac{1}{4}$, $-3$
  • (d)$\frac{1}{3}$, $4$
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(c) $\frac{1}{3}$, $-4$
211 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)$-1,-3$
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(d) $-1, -3$
221 Mark · March 2023 · Standardopen ↗
The zeroes of the polynomial $3x^2 + 11x - 4$ are :
  • (a)$\frac{1}{3}$, $-4$
  • (b)$-\frac{1}{3}$, $4$
  • (c)$\frac{1}{3}$, $4$
  • (d)$-\frac{1}{3}$, $-4$
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(A) $\frac{1}{3}$, $-4$
231 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)$3,\frac{2}{2}$
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(A) $3, -\frac{3}{2}$
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)$-3, -\frac{3}{2}$
  • (d)$3, \frac{3}{2}$
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(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
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(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)$\sqrt{2},2$
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Sol. (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)$\frac{2}{3}, -\frac{1}{7}$
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(D) $\frac{2}{3}, -\frac{1}{7}$
281 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)$\sqrt{2}, 2$
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Sol. (B) $2\sqrt{2}, \sqrt{2}$
291 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)$-3$
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(D) $-3$
2 Marks Questions
302 Marks · March 2025 · Standardopen ↗
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 $\frac{2}{3}, -2$.
312 Marks · March 2025 · Standardopen ↗
If $p$ and $q$ are zeroes of the polynomial $p(y) = 21y^2 - y - 2$, then find the value of $(1-p) . (1 - q)$.
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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 $\frac{6}{7}$

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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Ans. (b) $\frac{37}{4}$
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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Ans. (a) $-\frac{2}{3}$
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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(A) 1
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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(B) $\frac{5}{4}$
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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(a) $-4$
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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(b) $a^2 + 2b$
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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(b) $2$
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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(c) $0$
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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Sol. (d) $0$
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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Sol. (d) $-\frac{3}{7}$
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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(d) $-\frac{3}{7}$
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
  • (a)2
  • (b)-1
  • (c)1
  • (d)0
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(D) 0
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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Sol. (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)$\frac{1}{2}$
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Sol. (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)$10$
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Sol. (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)$\frac{5}{24}$
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(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)$\frac{3}{-7}$
  • (c)$\frac{3}{5}$
  • (d)$\frac{5}{-7}$
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(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)$1$
  • (c)$\frac{61}{4}$
  • (d)$\frac{71}{4}$
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(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
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(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)$21:5$
  • (c)$2:7$
  • (d)$7:2$
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(B) $2:7$
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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(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)$-\frac{1}{2}$
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(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)$4$
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(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)$-\frac{5}{7}$
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(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)$2$
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(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)$-a+b$
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Sol. (C) $a - b$
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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Sol. (D)$\frac{2}{3}$
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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(A) 2
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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(B) $\pm 15$
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 $(x^2 - 3x - 4)$

Making polynomials from zeroes

1 Mark Questions
921 Mark · March 2023 · Standardopen ↗
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) $5 (9x^2 – 4)$
931 Mark · March 2023 · Standardopen ↗
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) $3x (3x - 5)$
941 Mark · March 2023 · Standardopen ↗
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
951 Mark · March 2023 · Standardopen ↗
The number of polynomials having zeroes $-1$ and $2$ is :
  • (a)exactly $2$
  • (b)at most $2$
  • (c)only $1$
  • (d)infinite
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(d) infinite
961 Mark · March 2023 · Standardopen ↗
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
971 Mark · March 2024 · Standardopen ↗
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) $x^2 - 4x - 1$
981 Mark · July 2025 · Standardopen ↗
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) $x^2 - 3x$
2 Marks Questions
992 Marks · March 2025 · Standardopen ↗
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)
3 Marks Questions
1003 Marks · March 2024 · Standardopen ↗
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 $x^2 - 8x + 15$

Real life application

4 Marks Questions
1014 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
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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 $\alpha \beta = 45$
$(\alpha - \beta)^2 = 144$
$\Rightarrow (\alpha + \beta)^2 - 4\alpha\beta = 144$
$\Rightarrow (-p)^2 - 4 \times 45 = 144$
$\Rightarrow p = \pm 18$
1024 Marks · March 2023 · Standardopen ↗
In a pool at an aquarium, a dolphin jumps out of the water travelling at $20$ cm per second. Its height above water level after $t$ seconds is given by $h = 20t - 16t^2$.
Based on the above, answer the following questions :
(i) Find zeroes of polynomial $p(t) = 20t - 16t^2$.
(ii) Which of the following types of graph represents $p(t)$?
(iii) (a) What would be the value of $h$ at $t = \frac{3}{2}$ ? Interpret the result.
OR
(iii) (b) How much distance has the dolphin covered before hitting the water level again?
figure for this question
figure for this question
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(i) $-16t^2 + 20t = 0 \Rightarrow 4t(-4t + 5) = 0$
$t = 0, t = \frac{5}{4}$
(ii) (a)
(iii)(a) At $t = \frac{3}{2}$, $h = -16 \times \frac{9}{4} + 20 \times \frac{3}{2} = -36 + 30 = -6$
It means after $\frac{3}{2}$ seconds, dolphin has reached $6$ cm below water level.
OR
(iii)(b) Speed of dolphin $= 20$ cm per second.
In one second, distance covered $= 20$ cm
In $\frac{5}{4}$ seconds, distance covered $= 20 \times \frac{5}{4} = 25$ cm
1034 Marks · March 2024 · Standardopen ↗
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.
figure for this question
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(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) $-5t^2 + 25t = 20$
$\Rightarrow t^2 - 5t + 4 = 0$
$\Rightarrow (t - 4)(t - 1) = 0$
$\Rightarrow t = 4, 1$

General

1 Mark Questions
1041 Mark · March 2023 · Standardopen ↗
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.
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(d) Assertion (A) is false, but Reason (R) is true.