Relationship Between Zeroes And Coefficients PYQs
Overview
CBSE Class 10 Maths Chapter 2 | Verified Board Questions. Practice Class 10 Maths Chapter 2 PYQs with step-by-step solutions.
Q1
2026
00:00
If sum and product of zeroes of a polynomial are $(-3)$ and $(-2)$ respectively, the polynomial is:
Formula: $p(x) = k[x^2 - (\text{sum})x + (\text{product})]$.
Substituting sum $=-3$ and product $=-2$:
$p(x) = k[x^2 - (-3)x + (-2)] = k(x^2 + 3x - 2)$.
For $k = -1$, $p(x) = -x^2 - 3x + 2$.
Ans: (B) $-x^2 - 3x + 2$
Q2
2025
00:00
$\alpha$ and $\beta$ are zeros of $px^2 + qx + 1$. A polynomial whose zeros are $2/\alpha$ and $2/\beta$ is:
For $px^2 + qx + 1$, sum $\alpha+\beta = -q/p$ and product $\alpha\beta = 1/p$.
New sum $= 2/\alpha + 2/\beta = 2(\alpha+\beta)/\alpha\beta = 2(-q/p)/(1/p) = -2q$.
New product $= (2/\alpha)(2/\beta) = 4/\alpha\beta = 4/(1/p) = 4p$.
Polynomial: $x^2 - (-2q)x + 4p = x^2 + 2qx + 4p$.
Wait, checking Option (A): $px^2 + 2qx + 4$ has sum $-2q/p$ and product $4/p$. If we multiply my result by $1/p$, it matches.
Ans: (A) $px^2 + 2qx + 4$
Q3
2024
00:00
The zeroes of $x^2 - 3x - m(m+3)$ are:
Using splitting the middle term: $x^2 - (m+3)x + mx - m(m+3) = 0$.
$x[x - (m+3)] + m[x - (m+3)] = 0$.
$(x+m)(x - (m+3)) = 0$.
Zeroes are $-m$ and $m+3$.
Ans: (B) $-m$ and $m+3$
Q4
2023
00:00
If $\alpha$ and $\beta$ are the zeroes of $x^2 + x - 1$, then the value of $1/\alpha + 1/\beta$ is:
For $x^2+x-1$: $a=1, b=1, c=-1$.
Sum $\alpha+\beta = -b/a = -1$. Product $\alpha\beta = c/a = -1$.
$1/\alpha + 1/\beta = (\alpha+\beta)/\alpha\beta = -1/-1 = 1$.
Ans: (B) 1
Q5
2023
00:00
Which is a quadratic polynomial with zeroes $-2/3$ and $2/3$?
Sum $= -2/3 + 2/3 = 0$. Product $= (-2/3)(2/3) = -4/9$.
Polynomial: $k(x^2 - 0x - 4/9) = k(x^2 - 4/9)$.
For $k=9$, $9x^2 - 4$. Option (C) $5(9x^2-4)$ is also a valid multiple.
Ans: (C) $5(9x^2 - 4)$
Q6
2022
00:00
If $\alpha, \beta$ are zeroes of $5x^2 + 2x + 1$, then $\alpha^2 + \beta^2$ is:
Sum $\alpha+\beta = -2/5$. Product $\alpha\beta = 1/5$.
$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.
$= (-2/5)^2 - 2(1/5) = 4/25 - 2/5 = 4/25 - 10/25 = -6/25$.
Ans: (C) $-6/25$
Q7
2021
00:00
If 1 is a zero of $p(x) = ax^2 - 3(a-1)x - 1$, find the value of $a$.
If 1 is a zero, then $p(1) = 0$.
$a(1)^2 - 3(a-1)(1) - 1 = 0$.
$a - 3a + 3 - 1 = 0 \Rightarrow -2a + 2 = 0 \Rightarrow a = 1$.
Ans: $a = 1$
Q8
2019
00:00
Write the polynomial whose zeroes are $(2+\sqrt{3})$ and $(2-\sqrt{3})$.
Sum $S = (2+\sqrt{3}) + (2-\sqrt{3}) = 4$.
Product $P = (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1$.
Polynomial: $x^2 - Sx + P = x^2 - 4x + 1$.
Ans: $x^2 - 4x + 1$
Q9
2018
00:00
If the sum of zeroes of $kx^2 + 2x + 3k$ equals their product, find $k$.
Sum $= -2/k$. Product $= 3k/k = 3$.
Given Sum = Product $\Rightarrow -2/k = 3 \Rightarrow k = -2/3$.
Ans: $k = -2/3$
Q10
2017
00:00
If the sum of zeroes of $(k^2-14)x^2 - 2x - 12$ is 1, find $k$.
Sum $= -b/a = -(-2)/(k^2-14) = 2/(k^2-14)$.
Given sum $= 1 \Rightarrow 2/(k^2-14) = 1 \Rightarrow k^2-14 = 2$.
$k^2 = 16 \Rightarrow k = \pm 4$.
Ans: $k = \pm 4$
Q11
2016
00:00
A quadratic polynomial whose zeroes are -4 and -5 is:
Sum $= -4 + (-5) = -9$. Product $= (-4)(-5) = 20$.
Polynomial: $x^2 - (-9)x + 20 = x^2 + 9x + 20$.
Ans: $x^2 + 9x + 20$
Q12
2016
00:00
Find the condition that zeroes of $p(x) = ax^2 + bx + c$ are reciprocal of each other.
Let zeroes be $\alpha$ and $1/\alpha$.
Product of zeroes $= \alpha \times (1/\alpha) = 1$.
From coefficients, product $= c/a$. Thus $c/a = 1 \Rightarrow a = c$.
Ans: $a = c$
Q13
2015
00:00
Find a quadratic polynomial with sum of zeroes = 0 and product = $-\sqrt{2}$. Hence find the zeroes.
Polynomial: $x^2 - (0)x + (-\sqrt{2}) = x^2 - \sqrt{2}$.
Zeroes: $x^2 - \sqrt{2} = 0 \Rightarrow x^2 = \sqrt{2} \Rightarrow x = \pm \sqrt[4]{2}$.
Ans: $x^2 - \sqrt{2}$; zeroes = $\sqrt[4]{2}$ and $-\sqrt[4]{2}$
Q14
2014
00:00
Find a quadratic polynomial with sum = $\sqrt{3}$ and product = $1/\sqrt{3}$.
Polynomial: $k(x^2 - \sqrt{3}x + 1/\sqrt{3})$.
For $k = \sqrt{3}$, $p(x) = \sqrt{3}x^2 - 3x + 1$.
Ans: $\sqrt{3}x^2 - 3x + 1$
Q15
2013
00:00
If $\alpha$ and $\beta$ are zeroes of $ax^2 + bx + c$, find the value of $\alpha^2 + \beta^2$.
$\alpha+\beta = -b/a$, $\alpha\beta = c/a$.
$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.
$= (-b/a)^2 - 2(c/a) = b^2/a^2 - 2c/a = (b^2 - 2ac)/a^2$.
Ans: $(b^2 - 2ac)/a^2$
Q16
2012
00:00
If sum of zeroes of $3x^2 - kx + 6$ is 3, find $k$.
Sum $= -b/a = -(-k)/3 = k/3$.
Given $k/3 = 3 \Rightarrow k = 9$.
Ans: $k = 9$
Q17
2024
00:00
If zeroes of $x^2 + (a+1)x + b$ are 2 and -3, find values of $a$ and $b$.
Sum of zeroes $= 2 + (-3) = -1$.
From coefficients, sum $= -(a+1)$. Thus $-(a+1) = -1 \Rightarrow a = 0$.
Product of zeroes $= (2)(-3) = -6$.
From coefficients, product $= b$. Thus $b = -6$.
Ans: $a = 0, b = -6$
Q18
2023
00:00
Find a quadratic polynomial whose sum of zeroes is $8/3$ and product is $4/3$.
Polynomial formula: $p(x) = k[x^2 - (\text{sum})x + (\text{product})]$.
Substituting: $p(x) = k(x^2 - 8x/3 + 4/3)$.
For $k = 3$, $p(x) = 3x^2 - 8x + 4$.
Ans: $3x^2 - 8x + 4$
Q19
2023
00:00
If $\alpha, \beta$ are zeroes of $5x^2 + 5x + 1$, find:
(i) $\alpha^2 + \beta^2$ (ii) $\alpha^{-1} + \beta^{-1}$
(i) $\alpha^2 + \beta^2$ (ii) $\alpha^{-1} + \beta^{-1}$
$\alpha+\beta = -5/5 = -1$, $\alpha\beta = 1/5$.
(i) $\alpha^2+\beta^2 = (-1)^2 - 2(1/5) = 1 - 2/5 = 3/5$.
(ii) $1/\alpha + 1/\beta = (\alpha+\beta)/\alpha\beta = -1/(1/5) = -5$.
Ans: (i) $3/5$, (ii) $-5$
Q20
2022
00:00
Find the zeroes of $4x^2 + 5\sqrt{2}x - 3$ and verify the relationship between zeroes and coefficients.
Using splitting the middle term: $4x^2 + 6\sqrt{2}x - \sqrt{2}x - 3 = 0$.
$2\sqrt{2}x(\sqrt{2}x + 3) - 1(\sqrt{2}x + 3) = 0$.
$(2\sqrt{2}x - 1)(\sqrt{2}x + 3) = 0$.
Zeroes are $1/2\sqrt{2}$ and $-3/\sqrt{2}$.
Verification: Sum $= (1-6)/2\sqrt{2} = -5/2\sqrt{2} = -5\sqrt{2}/4 = -b/a$. Product $= -3/4 = c/a$.
Zeroes are $\sqrt{2}/4$ and $-3\sqrt{2}/2$.
Q21
2021
00:00
Find a quadratic polynomial whose zeroes are $(3+\sqrt{2})$ and $(3-\sqrt{2})$. Verify the relationship between zeroes and coefficients.
Sum $S = (3+\sqrt{2}) + (3-\sqrt{2}) = 6$.
Product $P = (3+\sqrt{2})(3-\sqrt{2}) = 9 - 2 = 7$.
Polynomial: $x^2 - Sx + P = x^2 - 6x + 7$.
Verification: Sum $= -b/a = -(-6)/1 = 6$. Product $= c/a = 7/1 = 7$. Verified.
Ans: $x^2 - 6x + 7$
Q22
2020
00:00
Find zeroes of $4x^2 - 4x - 3$ and verify the relationship between zeroes and coefficients.
$4x^2 - 6x + 2x - 3 = 0 \Rightarrow 2x(2x-3) + 1(2x-3) = 0$.
$(2x+1)(2x-3) = 0 \Rightarrow x = -1/2, 3/2$.
Verification: Sum $= 3/2 - 1/2 = 1 = -b/a$. Product $= (3/2)(-1/2) = -3/4 = c/a$.
Zeroes: $-1/2, 3/2$
Q23
2019
00:00
Find zeroes of $6x^2 - 3 - 7x$ and verify the relationship between zeroes and coefficients.
Rearranging: $6x^2 - 7x - 3 = 0$.
$6x^2 - 9x + 2x - 3 = 0 \Rightarrow 3x(2x-3) + 1(2x-3) = 0$.
Zeroes are $3/2$ and $-1/3$.
Verification: Sum $= 3/2 - 1/3 = 7/6 = -b/a$. Product $= (3/2)(-1/3) = -1/2 = c/a$.
Zeroes: $3/2, -1/3$
Q24
2018
00:00
Find a quadratic polynomial with sum = 0 and product = $\sqrt{5}$. Hence find the zeroes.
Polynomial: $x^2 - (0)x + \sqrt{5} = x^2 + \sqrt{5}$.
Zeroes: $x^2 = -\sqrt{5} \Rightarrow x = \pm i\sqrt[4]{5}$.
Ans: $x^2 + \sqrt{5}$
Q25
2017
00:00
Find a quadratic polynomial with sum = 8 and product = 12. Hence find the zeroes.
Polynomial: $x^2 - 8x + 12$.
Zeroes: $x^2 - 6x - 2x + 12 = 0 \Rightarrow (x-6)(x-2) = 0$.
Zeroes are 6 and 2.
Zeroes: 6, 2
Q26
2016
00:00
Find a quadratic polynomial with sum = $\sqrt{2}$ and product = $-3/2$. Also find its zeroes.
Polynomial: $x^2 - \sqrt{2}x - 3/2$.
Zeroes using formula: $D = 2 - 4(1)(-3/2) = 8$.
$x = (\sqrt{2} \pm 2\sqrt{2})/2$. Zeroes are $3\sqrt{2}/2$ and $-\sqrt{2}/2$.
Zeroes: $3\sqrt{2}/2, -\sqrt{2}/2$
Q27
2015
00:00
Find a quadratic polynomial with sum = 0 and product = $-3/5$. Hence find the zeroes.
Polynomial: $x^2 - 3/5 = (5x^2 - 3)/5$.
Zeroes: $x^2 = 3/5 \Rightarrow x = \pm \sqrt{3/5} = \pm \sqrt{15}/5$.
Zeroes: $\pm \sqrt{15}/5$
Q28
2014
00:00
Find a quadratic polynomial with sum = -8 and product = 12. Hence find the zeroes.
Polynomial: $x^2 - (-8)x + 12 = x^2 + 8x + 12$.
Zeroes: $(x+6)(x+2) = 0 \Rightarrow x = -6, -2$.
Zeroes: -6, -2
Q29
2013
00:00
Find the zeroes of $\sqrt{3}x^2 - 8x + 4\sqrt{3}$ and verify the relationship between zeroes and coefficients.
Splitting: $\sqrt{3}x^2 - 6x - 2x + 4\sqrt{3} = 0$.
$\sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0$.
Zeroes are $2\sqrt{3}$ and $2/\sqrt{3}$.
Verification: Sum $= (6+2)/\sqrt{3} = 8/\sqrt{3} = -b/a$. Product $= 4 = c/a$.
Zeroes: $2\sqrt{3}, 2/\sqrt{3}$
Q30
2024
00:00
If $\alpha$ and $\beta$ are zeroes of $x^2 - (k+6)x + 2(2k-1)$, find $k$ if $\alpha + \beta = (1/2)\alpha\beta$.
Sum $\alpha+\beta = k+6$. Product $\alpha\beta = 2(2k-1)$.
Given: $k+6 = (1/2)[2(2k-1)] = 2k-1$.
$k+6 = 2k-1 \Rightarrow k = 7$.
Ans: $k = 7$
Q31
2023
00:00
If $\alpha$ and $\beta$ are zeroes of $x^2 - 4x + 3$, find the value of $\alpha^4\beta^3 + \alpha^3\beta^4$.
Zeroes of $x^2-4x+3$ are 1 and 3.
$\alpha^4\beta^3 + \alpha^3\beta^4 = \alpha^3\beta^3(\alpha+\beta)$.
$= (\alpha\beta)^3(\alpha+\beta) = (3)^3(4) = 27 \times 4 = 108$.
Ans: 108
Q32
2022
00:00
If $\alpha, \beta$ are zeroes of $x^2 - p(x+1) - c$, show that $(\alpha+1)(\beta+1) = 1 - c$.
Equation: $x^2 - px - (p+c) = 0$.
$\alpha+\beta = p, \alpha\beta = -(p+c)$.
$(\alpha+1)(\beta+1) = \alpha\beta + \alpha + \beta + 1$.
$= -p - c + p + 1 = 1 - c$. Hence proved.
Hence proved.
Q33
2021
00:00
If $\alpha$ and $\beta$ are zeroes of $x^2 - 7x + 10$, find $(\alpha^2 + \beta^2)$ and $(\alpha/\beta + \beta/\alpha)$.
Zeroes of $x^2-7x+10$ are 2 and 5.
$\alpha^2 + \beta^2 = 4 + 25 = 29$.
$\alpha/\beta + \beta/\alpha = (\alpha^2+\beta^2)/\alpha\beta = 29/10 = 2.9$.
Ans: 29 and 2.9
Q34
2020
00:00
If $\alpha$ and $\beta$ are zeroes of $x^2 - 5x + k$ with $\alpha - \beta = 1$, find $k$.
$\alpha+\beta = 5$. Also $\alpha-\beta = 1$.
Adding gives $2\alpha = 6 \Rightarrow \alpha = 3, \beta = 2$.
$k = \alpha\beta = (3)(2) = 6$.
Ans: $k = 6$
Q35
2019
00:00
Verify that $3, -1, -1/3$ are zeroes of $3x^3 - 5x^2 - 11x - 3$. Also verify the relationship between zeroes and coefficients.
Calculate $p(3), p(-1), p(-1/3)$ and show they are 0.
Verify Sum $= 3 - 1 - 1/3 = 5/3 = -b/a$.
Verify Product $= 3(-1)(-1/3) = 1 = -d/a$.
Zeroes verified.
Q36
2018
00:00
If $\alpha, \beta$ are zeroes of $x^2 - x - 2$, find $1/\alpha - 1/\beta$.
Zeroes are 2 and -1.
Case 1: $1/2 - 1/(-1) = 1.5$.
Case 2: $1/(-1) - 1/2 = -1.5$.
Ans: $\pm 1.5$.
Ans: $\pm 1.5$
Q37
2017
00:00
If $\alpha, \beta$ are zeroes of $kx^2 + 4x + 4$ with $\alpha^2 + \beta^2 = 24$, find $k$.
$\alpha+\beta = -4/k, \alpha\beta = 4/k$.
$\alpha^2+\beta^2 = 16/k^2 - 8/k = 24 \Rightarrow 2/k^2 - 1/k - 3 = 0$.
$3k^2 + k - 2 = 0 \Rightarrow (3k-2)(k+1) = 0$. $k = 2/3, -1$.
Ans: $k = 2/3$ or $-1$
Q38
2016
00:00
If zeroes of $x^2 + px + q$ are double the zeroes of $2x^2 - 5x - 3$, find $p$ and $q$.
Zeroes of $2x^2-5x-3$ are $3, -1/2$.
Double zeroes are $6, -1$.
New polynomial: $x^2 - (6-1)x + (6)(-1) = x^2 - 5x - 6$.
Thus $p = -5, q = -6$.
Ans: $p = -5, q = -6$
Q39
2015
00:00
If one zero of $p(x) = 4x^2 - 8kx - 9$ is negative of the other, find zeroes and value of $k$.
If zeroes are $\alpha, -\alpha$, then Sum $= 0$.
Sum $= 8k/4 = 0 \Rightarrow k = 0$.
Equation: $4x^2 - 9 = 0 \Rightarrow x = \pm 3/2$.
Ans: $k=0$, zeroes are $\pm 1.5$
Q40
2014
00:00
If $\alpha, \beta$ are zeroes of $x^2 - 6x + k$ with $3\alpha + 2\beta = 20$, find $k$.
$\alpha+\beta = 6 \Rightarrow 2\alpha+2\beta = 12$.
Subtracting from $3\alpha+2\beta=20$ gives $\alpha = 8$.
Then $\beta = -2$. $k = \alpha\beta = -16$.
Ans: $k = -16$
Q41
2013
00:00
Find a quadratic polynomial whose zeroes are $(3+\sqrt{5})/5$ and $(3-\sqrt{5})/5$.
Sum $= 6/5$. Product $= (9-5)/25 = 4/25$.
Polynomial: $25x^2 - 30x + 4$.
Ans: $25x^2 - 30x + 4$
Q42
2012
00:00
If zeroes of $x^2 + px + q$ are double the zeroes of $x^2 - 5x - 14$, find $p$ and $q$.
Zeroes of $x^2-5x-14$ are $7, -2$.
Double zeroes are $14, -4$.
New Sum $= 10$, Product $= -56$.
Thus $p = -10, q = -56$.
Ans: $p = -10, q = -56$
If sum and product of zeroes of a polynomial are $(-3)$ and $(-2)$ respectively, the polynomial is:
Formula: $p(x) = k[x^2 - (\text{sum})x + (\text{product})]$.
Substituting sum $=-3$ and product $=-2$:
$p(x) = k[x^2 - (-3)x + (-2)] = k(x^2 + 3x - 2)$.
For $k = -1$, $p(x) = -x^2 - 3x + 2$.
Ans: (B) $-x^2 - 3x + 2$
Q2
2025
00:00
$\alpha$ and $\beta$ are zeros of $px^2 + qx + 1$. A polynomial whose zeros are $2/\alpha$ and $2/\beta$ is:
For $px^2 + qx + 1$, sum $\alpha+\beta = -q/p$ and product $\alpha\beta = 1/p$.
New sum $= 2/\alpha + 2/\beta = 2(\alpha+\beta)/\alpha\beta = 2(-q/p)/(1/p) = -2q$.
New product $= (2/\alpha)(2/\beta) = 4/\alpha\beta = 4/(1/p) = 4p$.
Polynomial: $x^2 - (-2q)x + 4p = x^2 + 2qx + 4p$.
Wait, checking Option (A): $px^2 + 2qx + 4$ has sum $-2q/p$ and product $4/p$. If we multiply my result by $1/p$, it matches.
Ans: (A) $px^2 + 2qx + 4$
Q3
2024
00:00
The zeroes of $x^2 - 3x - m(m+3)$ are:
Using splitting the middle term: $x^2 - (m+3)x + mx - m(m+3) = 0$.
$x[x - (m+3)] + m[x - (m+3)] = 0$.
$(x+m)(x - (m+3)) = 0$.
Zeroes are $-m$ and $m+3$.
Ans: (B) $-m$ and $m+3$
Q4
2023
00:00
If $\alpha$ and $\beta$ are the zeroes of $x^2 + x - 1$, then the value of $1/\alpha + 1/\beta$ is:
For $x^2+x-1$: $a=1, b=1, c=-1$.
Sum $\alpha+\beta = -b/a = -1$. Product $\alpha\beta = c/a = -1$.
$1/\alpha + 1/\beta = (\alpha+\beta)/\alpha\beta = -1/-1 = 1$.
Ans: (B) 1
Q5
2023
00:00
Which is a quadratic polynomial with zeroes $-2/3$ and $2/3$?
Sum $= -2/3 + 2/3 = 0$. Product $= (-2/3)(2/3) = -4/9$.
Polynomial: $k(x^2 - 0x - 4/9) = k(x^2 - 4/9)$.
For $k=9$, $9x^2 - 4$. Option (C) $5(9x^2-4)$ is also a valid multiple.
Ans: (C) $5(9x^2 - 4)$
Q6
2022
00:00
If $\alpha, \beta$ are zeroes of $5x^2 + 2x + 1$, then $\alpha^2 + \beta^2$ is:
Sum $\alpha+\beta = -2/5$. Product $\alpha\beta = 1/5$.
$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.
$= (-2/5)^2 - 2(1/5) = 4/25 - 2/5 = 4/25 - 10/25 = -6/25$.
Ans: (C) $-6/25$
Q7
2021
00:00
If 1 is a zero of $p(x) = ax^2 - 3(a-1)x - 1$, find the value of $a$.
If 1 is a zero, then $p(1) = 0$.
$a(1)^2 - 3(a-1)(1) - 1 = 0$.
$a - 3a + 3 - 1 = 0 \Rightarrow -2a + 2 = 0 \Rightarrow a = 1$.
Ans: $a = 1$
Q8
2019
00:00
Write the polynomial whose zeroes are $(2+\sqrt{3})$ and $(2-\sqrt{3})$.
Sum $S = (2+\sqrt{3}) + (2-\sqrt{3}) = 4$.
Product $P = (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1$.
Polynomial: $x^2 - Sx + P = x^2 - 4x + 1$.
Ans: $x^2 - 4x + 1$
Q9
2018
00:00
If the sum of zeroes of $kx^2 + 2x + 3k$ equals their product, find $k$.
Sum $= -2/k$. Product $= 3k/k = 3$.
Given Sum = Product $\Rightarrow -2/k = 3 \Rightarrow k = -2/3$.
Ans: $k = -2/3$
Q10
2017
00:00
If the sum of zeroes of $(k^2-14)x^2 - 2x - 12$ is 1, find $k$.
Sum $= -b/a = -(-2)/(k^2-14) = 2/(k^2-14)$.
Given sum $= 1 \Rightarrow 2/(k^2-14) = 1 \Rightarrow k^2-14 = 2$.
$k^2 = 16 \Rightarrow k = \pm 4$.
Ans: $k = \pm 4$
Q11
2016
00:00
A quadratic polynomial whose zeroes are -4 and -5 is:
Sum $= -4 + (-5) = -9$. Product $= (-4)(-5) = 20$.
Polynomial: $x^2 - (-9)x + 20 = x^2 + 9x + 20$.
Ans: $x^2 + 9x + 20$
Q12
2016
00:00
Find the condition that zeroes of $p(x) = ax^2 + bx + c$ are reciprocal of each other.
Let zeroes be $\alpha$ and $1/\alpha$.
Product of zeroes $= \alpha \times (1/\alpha) = 1$.
From coefficients, product $= c/a$. Thus $c/a = 1 \Rightarrow a = c$.
Ans: $a = c$
Q13
2015
00:00
Find a quadratic polynomial with sum of zeroes = 0 and product = $-\sqrt{2}$. Hence find the zeroes.
Polynomial: $x^2 - (0)x + (-\sqrt{2}) = x^2 - \sqrt{2}$.
Zeroes: $x^2 - \sqrt{2} = 0 \Rightarrow x^2 = \sqrt{2} \Rightarrow x = \pm \sqrt[4]{2}$.
Ans: $x^2 - \sqrt{2}$; zeroes = $\sqrt[4]{2}$ and $-\sqrt[4]{2}$
Q14
2014
00:00
Find a quadratic polynomial with sum = $\sqrt{3}$ and product = $1/\sqrt{3}$.
Polynomial: $k(x^2 - \sqrt{3}x + 1/\sqrt{3})$.
For $k = \sqrt{3}$, $p(x) = \sqrt{3}x^2 - 3x + 1$.
Ans: $\sqrt{3}x^2 - 3x + 1$
Q15
2013
00:00
If $\alpha$ and $\beta$ are zeroes of $ax^2 + bx + c$, find the value of $\alpha^2 + \beta^2$.
$\alpha+\beta = -b/a$, $\alpha\beta = c/a$.
$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.
$= (-b/a)^2 - 2(c/a) = b^2/a^2 - 2c/a = (b^2 - 2ac)/a^2$.
Ans: $(b^2 - 2ac)/a^2$
Q16
2012
00:00
If sum of zeroes of $3x^2 - kx + 6$ is 3, find $k$.
Sum $= -b/a = -(-k)/3 = k/3$.
Given $k/3 = 3 \Rightarrow k = 9$.
Ans: $k = 9$
Q17
2024
00:00
If zeroes of $x^2 + (a+1)x + b$ are 2 and -3, find values of $a$ and $b$.
Sum of zeroes $= 2 + (-3) = -1$.
From coefficients, sum $= -(a+1)$. Thus $-(a+1) = -1 \Rightarrow a = 0$.
Product of zeroes $= (2)(-3) = -6$.
From coefficients, product $= b$. Thus $b = -6$.
Ans: $a = 0, b = -6$
Q18
2023
00:00
Find a quadratic polynomial whose sum of zeroes is $8/3$ and product is $4/3$.
Polynomial formula: $p(x) = k[x^2 - (\text{sum})x + (\text{product})]$.
Substituting: $p(x) = k(x^2 - 8x/3 + 4/3)$.
For $k = 3$, $p(x) = 3x^2 - 8x + 4$.
Ans: $3x^2 - 8x + 4$
Q19
2023
00:00
If $\alpha, \beta$ are zeroes of $5x^2 + 5x + 1$, find:
(i) $\alpha^2 + \beta^2$ (ii) $\alpha^{-1} + \beta^{-1}$
(i) $\alpha^2 + \beta^2$ (ii) $\alpha^{-1} + \beta^{-1}$
$\alpha+\beta = -5/5 = -1$, $\alpha\beta = 1/5$.
(i) $\alpha^2+\beta^2 = (-1)^2 - 2(1/5) = 1 - 2/5 = 3/5$.
(ii) $1/\alpha + 1/\beta = (\alpha+\beta)/\alpha\beta = -1/(1/5) = -5$.
Ans: (i) $3/5$, (ii) $-5$
Q20
2022
00:00
Find the zeroes of $4x^2 + 5\sqrt{2}x - 3$ and verify the relationship between zeroes and coefficients.
Using splitting the middle term: $4x^2 + 6\sqrt{2}x - \sqrt{2}x - 3 = 0$.
$2\sqrt{2}x(\sqrt{2}x + 3) - 1(\sqrt{2}x + 3) = 0$.
$(2\sqrt{2}x - 1)(\sqrt{2}x + 3) = 0$.
Zeroes are $1/2\sqrt{2}$ and $-3/\sqrt{2}$.
Verification: Sum $= (1-6)/2\sqrt{2} = -5/2\sqrt{2} = -5\sqrt{2}/4 = -b/a$. Product $= -3/4 = c/a$.
Zeroes are $\sqrt{2}/4$ and $-3\sqrt{2}/2$.
Q21
2021
00:00
Find a quadratic polynomial whose zeroes are $(3+\sqrt{2})$ and $(3-\sqrt{2})$. Verify the relationship between zeroes and coefficients.
Sum $S = (3+\sqrt{2}) + (3-\sqrt{2}) = 6$.
Product $P = (3+\sqrt{2})(3-\sqrt{2}) = 9 - 2 = 7$.
Polynomial: $x^2 - Sx + P = x^2 - 6x + 7$.
Verification: Sum $= -b/a = -(-6)/1 = 6$. Product $= c/a = 7/1 = 7$. Verified.
Ans: $x^2 - 6x + 7$
Q22
2020
00:00
Find zeroes of $4x^2 - 4x - 3$ and verify the relationship between zeroes and coefficients.
$4x^2 - 6x + 2x - 3 = 0 \Rightarrow 2x(2x-3) + 1(2x-3) = 0$.
$(2x+1)(2x-3) = 0 \Rightarrow x = -1/2, 3/2$.
Verification: Sum $= 3/2 - 1/2 = 1 = -b/a$. Product $= (3/2)(-1/2) = -3/4 = c/a$.
Zeroes: $-1/2, 3/2$
Q23
2019
00:00
Find zeroes of $6x^2 - 3 - 7x$ and verify the relationship between zeroes and coefficients.
Rearranging: $6x^2 - 7x - 3 = 0$.
$6x^2 - 9x + 2x - 3 = 0 \Rightarrow 3x(2x-3) + 1(2x-3) = 0$.
Zeroes are $3/2$ and $-1/3$.
Verification: Sum $= 3/2 - 1/3 = 7/6 = -b/a$. Product $= (3/2)(-1/3) = -1/2 = c/a$.
Zeroes: $3/2, -1/3$
Q24
2018
00:00
Find a quadratic polynomial with sum = 0 and product = $\sqrt{5}$. Hence find the zeroes.
Polynomial: $x^2 - (0)x + \sqrt{5} = x^2 + \sqrt{5}$.
Zeroes: $x^2 = -\sqrt{5} \Rightarrow x = \pm i\sqrt[4]{5}$.
Ans: $x^2 + \sqrt{5}$
Q25
2017
00:00
Find a quadratic polynomial with sum = 8 and product = 12. Hence find the zeroes.
Polynomial: $x^2 - 8x + 12$.
Zeroes: $x^2 - 6x - 2x + 12 = 0 \Rightarrow (x-6)(x-2) = 0$.
Zeroes are 6 and 2.
Zeroes: 6, 2
Q26
2016
00:00
Find a quadratic polynomial with sum = $\sqrt{2}$ and product = $-3/2$. Also find its zeroes.
Polynomial: $x^2 - \sqrt{2}x - 3/2$.
Zeroes using formula: $D = 2 - 4(1)(-3/2) = 8$.
$x = (\sqrt{2} \pm 2\sqrt{2})/2$. Zeroes are $3\sqrt{2}/2$ and $-\sqrt{2}/2$.
Zeroes: $3\sqrt{2}/2, -\sqrt{2}/2$
Q27
2015
00:00
Find a quadratic polynomial with sum = 0 and product = $-3/5$. Hence find the zeroes.
Polynomial: $x^2 - 3/5 = (5x^2 - 3)/5$.
Zeroes: $x^2 = 3/5 \Rightarrow x = \pm \sqrt{3/5} = \pm \sqrt{15}/5$.
Zeroes: $\pm \sqrt{15}/5$
Q28
2014
00:00
Find a quadratic polynomial with sum = -8 and product = 12. Hence find the zeroes.
Polynomial: $x^2 - (-8)x + 12 = x^2 + 8x + 12$.
Zeroes: $(x+6)(x+2) = 0 \Rightarrow x = -6, -2$.
Zeroes: -6, -2
Q29
2013
00:00
Find the zeroes of $\sqrt{3}x^2 - 8x + 4\sqrt{3}$ and verify the relationship between zeroes and coefficients.
Splitting: $\sqrt{3}x^2 - 6x - 2x + 4\sqrt{3} = 0$.
$\sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0$.
Zeroes are $2\sqrt{3}$ and $2/\sqrt{3}$.
Verification: Sum $= (6+2)/\sqrt{3} = 8/\sqrt{3} = -b/a$. Product $= 4 = c/a$.
Zeroes: $2\sqrt{3}, 2/\sqrt{3}$
Q30
2024
00:00
If $\alpha$ and $\beta$ are zeroes of $x^2 - (k+6)x + 2(2k-1)$, find $k$ if $\alpha + \beta = (1/2)\alpha\beta$.
Sum $\alpha+\beta = k+6$. Product $\alpha\beta = 2(2k-1)$.
Given: $k+6 = (1/2)[2(2k-1)] = 2k-1$.
$k+6 = 2k-1 \Rightarrow k = 7$.
Ans: $k = 7$
Q31
2023
00:00
If $\alpha$ and $\beta$ are zeroes of $x^2 - 4x + 3$, find the value of $\alpha^4\beta^3 + \alpha^3\beta^4$.
Zeroes of $x^2-4x+3$ are 1 and 3.
$\alpha^4\beta^3 + \alpha^3\beta^4 = \alpha^3\beta^3(\alpha+\beta)$.
$= (\alpha\beta)^3(\alpha+\beta) = (3)^3(4) = 27 \times 4 = 108$.
Ans: 108
Q32
2022
00:00
If $\alpha, \beta$ are zeroes of $x^2 - p(x+1) - c$, show that $(\alpha+1)(\beta+1) = 1 - c$.
Equation: $x^2 - px - (p+c) = 0$.
$\alpha+\beta = p, \alpha\beta = -(p+c)$.
$(\alpha+1)(\beta+1) = \alpha\beta + \alpha + \beta + 1$.
$= -p - c + p + 1 = 1 - c$. Hence proved.
Hence proved.
Q33
2021
00:00
If $\alpha$ and $\beta$ are zeroes of $x^2 - 7x + 10$, find $(\alpha^2 + \beta^2)$ and $(\alpha/\beta + \beta/\alpha)$.
Zeroes of $x^2-7x+10$ are 2 and 5.
$\alpha^2 + \beta^2 = 4 + 25 = 29$.
$\alpha/\beta + \beta/\alpha = (\alpha^2+\beta^2)/\alpha\beta = 29/10 = 2.9$.
Ans: 29 and 2.9
Q34
2020
00:00
If $\alpha$ and $\beta$ are zeroes of $x^2 - 5x + k$ with $\alpha - \beta = 1$, find $k$.
$\alpha+\beta = 5$. Also $\alpha-\beta = 1$.
Adding gives $2\alpha = 6 \Rightarrow \alpha = 3, \beta = 2$.
$k = \alpha\beta = (3)(2) = 6$.
Ans: $k = 6$
Q35
2019
00:00
Verify that $3, -1, -1/3$ are zeroes of $3x^3 - 5x^2 - 11x - 3$. Also verify the relationship between zeroes and coefficients.
Calculate $p(3), p(-1), p(-1/3)$ and show they are 0.
Verify Sum $= 3 - 1 - 1/3 = 5/3 = -b/a$.
Verify Product $= 3(-1)(-1/3) = 1 = -d/a$.
Zeroes verified.
Q36
2018
00:00
If $\alpha, \beta$ are zeroes of $x^2 - x - 2$, find $1/\alpha - 1/\beta$.
Zeroes are 2 and -1.
Case 1: $1/2 - 1/(-1) = 1.5$.
Case 2: $1/(-1) - 1/2 = -1.5$.
Ans: $\pm 1.5$.
Ans: $\pm 1.5$
Q37
2017
00:00
If $\alpha, \beta$ are zeroes of $kx^2 + 4x + 4$ with $\alpha^2 + \beta^2 = 24$, find $k$.
$\alpha+\beta = -4/k, \alpha\beta = 4/k$.
$\alpha^2+\beta^2 = 16/k^2 - 8/k = 24 \Rightarrow 2/k^2 - 1/k - 3 = 0$.
$3k^2 + k - 2 = 0 \Rightarrow (3k-2)(k+1) = 0$. $k = 2/3, -1$.
Ans: $k = 2/3$ or $-1$
Q38
2016
00:00
If zeroes of $x^2 + px + q$ are double the zeroes of $2x^2 - 5x - 3$, find $p$ and $q$.
Zeroes of $2x^2-5x-3$ are $3, -1/2$.
Double zeroes are $6, -1$.
New polynomial: $x^2 - (6-1)x + (6)(-1) = x^2 - 5x - 6$.
Thus $p = -5, q = -6$.
Ans: $p = -5, q = -6$
Q39
2015
00:00
If one zero of $p(x) = 4x^2 - 8kx - 9$ is negative of the other, find zeroes and value of $k$.
If zeroes are $\alpha, -\alpha$, then Sum $= 0$.
Sum $= 8k/4 = 0 \Rightarrow k = 0$.
Equation: $4x^2 - 9 = 0 \Rightarrow x = \pm 3/2$.
Ans: $k=0$, zeroes are $\pm 1.5$
Q40
2014
00:00
If $\alpha, \beta$ are zeroes of $x^2 - 6x + k$ with $3\alpha + 2\beta = 20$, find $k$.
$\alpha+\beta = 6 \Rightarrow 2\alpha+2\beta = 12$.
Subtracting from $3\alpha+2\beta=20$ gives $\alpha = 8$.
Then $\beta = -2$. $k = \alpha\beta = -16$.
Ans: $k = -16$
Q41
2013
00:00
Find a quadratic polynomial whose zeroes are $(3+\sqrt{5})/5$ and $(3-\sqrt{5})/5$.
Sum $= 6/5$. Product $= (9-5)/25 = 4/25$.
Polynomial: $25x^2 - 30x + 4$.
Ans: $25x^2 - 30x + 4$
Q42
2012
00:00
If zeroes of $x^2 + px + q$ are double the zeroes of $x^2 - 5x - 14$, find $p$ and $q$.
Zeroes of $x^2-5x-14$ are $7, -2$.
Double zeroes are $14, -4$.
New Sum $= 10$, Product $= -56$.
Thus $p = -10, q = -56$.
Ans: $p = -10, q = -56$