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Nature Of Roots PYQs

Overview

This page provides comprehensive Class 10 Maths Nature Of Roots PYQs | Quadratic Equations. Nature Of Roots previous year questions for Class 10 Maths Quadratic Equations. Practice CBSE board PYQs with step-by-step solutions on SJMaths.

Practice Previous Year Questions Topic-wise

Q1 2024
00:00
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
Sum of roots: $-\frac{b}{a}=\frac{6}{5}$
Product of roots: $\frac{c}{a}=\frac{21}{5}$
Ratio: $\frac{6}{5}:\frac{21}{5}=6:21=2:7$
Final Answer: (b) 2 : 7
Q2 2023
00:00
Which of the following quadratic equations has sum of its roots as 4?
(a)$2x^2-4x+8=0$
(b)$-x^2+4x+4=0$
(c)$\sqrt{2}x^2-\frac{4}{\sqrt{2}}x+1=0$
(d)$4x^2-4x+4=0$
Formula: Sum of roots $=-\frac{b}{a}$
For option (b): $-\frac{4}{-1}=4$
Final Answer: (b)
Q3 2025
00:00
The discriminant of the quadratic equation $bx^2+ax+c=0$ is:
(a)$b^2-4ac$
(b)$\sqrt{b^2-4ac}$
(c)$\sqrt{a^2-4bc}$
(d)$a^2-4bc$
Discriminant: $D=b^2-4ac$
Here coefficient of $x$ is $a$
$D=a^2-4bc$
Final Answer: (d)
Q4 2020
00:00
The discriminant of $3\sqrt{3}x^2+10x+\sqrt{3}=0$ is:
(a)±8
(b)8
(c)$100-4\sqrt{3}$
(d)64
$D=b^2-4ac$
$=100-4(3\sqrt{3})(\sqrt{3})$
$=100-36=64$
Final Answer: (d) 64
Q5 2025
00:00
Which equation has real and equal roots?
(a)$(x+1)^2=2x+1$
(b)$x^2+x=0$
(c)$x^2-4=0$
(d)$x^2+x+1=0$
$(x+1)^2=2x+1 \Rightarrow x^2=0$
Discriminant $=0$
Final Answer: (a)
Q6 2024
00:00
If the roots of $ax^2+bx+c=0$ are real and equal, which relation is true?
(a)$a=\frac{b^2}{c}$
(b)$b^2=ac$
(c)$ac=\frac{b^2}{4}$
(d)$c=\frac{b^2}{a}$
$D=b^2-4ac=0$
$ac=\frac{b^2}{4}$
Final Answer: (c)
Q7 2023
00:00
Values of $k$ for which $4x^2+kx+9=0$ has equal roots:
(a)±11
(b)±12
(c)±6
(d)±3
$D=k^2-144=0$
$k=±12$
Final Answer: (b)
Q8 2023
00:00
Least positive value of $k$ for which $2x^2+kx-4=0$ has rational roots:
(a)±$2\sqrt{2}$
(b)2
(c)±2
(d)$\sqrt{2}$
$D=k^2+32$ must be perfect square
Least such $k=2$
Final Answer: (b)
Q9 2020
00:00
Value(s) of $k$ for which $2x^2+kx+2=0$ has equal roots:
(a)4
(b)±4
(c)−4
(d)0
$D=k^2-16=0$
$k=±4$
Final Answer: (b)
Q10 2019
00:00
Write the discriminant of the quadratic equation $(x+5)^2=2(5x-3)$.
Given: $(x+5)^2=10x-6$
$x^2+10x+25-10x+6=0$
$x^2+31=0$
$D=b^2-4ac=0-124=-124$
Final Answer: Discriminant = −124
Q11 VSA · 2019
00:00
Write the discriminant of the quadratic equation $(x + 5)^2 = 2(5x − 3)$.
$ (x+5)^2 = 10x - 6 $
$ x^2 + 10x + 25 - 10x + 6 = 0 $
$ x^2 + 31 = 0 $
Here, $a=1,\; b=0,\; c=31$
Discriminant $D=b^2-4ac=0-124=-124$
Final Answer: Discriminant = −124
Q12 VSA · 2019
00:00
Find the nature of roots of the quadratic equation $2x^2 − 4x + 3 = 0$.
$a=2,\; b=-4,\; c=3$
$D=b^2-4ac=16-24=-8$
Since $D<0$, the roots are non-real (complex).
Q13 SA-I · 2023
00:00
Find the sum and product of the roots of the quadratic equation $2x^2 − 9x + 4 = 0$.
$a=2,\; b=-9,\; c=4$
Sum of roots $= -\frac{b}{a}=\frac{9}{2}$
Product of roots $=\frac{c}{a}=\frac{4}{2}=2$
Final Answer: Sum = 9/2, Product = 2
Q14 SA-I · 2017
00:00
Find the value of $p$ for which one root of $px^2 − 14x + 8 = 0$ is six times the other.
Let roots be $\alpha$ and $6\alpha$
Sum: $\alpha+6\alpha=7\alpha=\frac{14}{p}$
$\alpha=\frac{2}{p}$
Product: $6\alpha^2=\frac{8}{p}$
$6\left(\frac{4}{p^2}\right)=\frac{8}{p}$
$p=3$
Final Answer: $p=3$
Q15 SA-I · 2019
00:00
If one root of the quadratic equation $2x^2+2x+k=0$ is $-1/3$, find the value of $k$.
Substitute $x=-1/3$
$2\left(\frac{1}{9}\right)-\frac{2}{3}+k=0$
$\frac{2}{9}-\frac{6}{9}+k=0$
$k=\frac{4}{9}$
Final Answer: $k=4/9$
Q16 SA-I · 2019
00:00
Find the value of $k$ for which the roots of $3x^2 − 10x + k = 0$ are reciprocal of each other.
For reciprocal roots: $\frac{c}{a}=1$
$\frac{k}{3}=1$
$k=3$
Final Answer: $k=3$
Q17 SA-I · 2018
00:00
If $x=3$ is one root of $x^2 − 2kx − 6 = 0$, find the value of $k$.
Substitute $x=3$
$9-6k-6=0$
$3=6k$
$k=\frac{1}{2}$
Final Answer: $k=1/2$
Q18 SA-I · 2019
00:00
For what values of $a$ does the quadratic equation $9x^2 − 3ax + 1 = 0$ have equal roots?
Equal roots ⇒ $D=0$
$(-3a)^2-4(9)(1)=0$
$9a^2=36$
$a=±2$
Final Answer: $a=±2$
Q19 SA-II · 2021
00:00
One root of the quadratic equation $2x^2 - 8x - k = 0$ is $\frac{5}{2}$. Find the value of $k$ and the other root.
Given one root $x=\frac{5}{2}$
Substitute in equation:
$2\left(\frac{25}{4}\right)-8\left(\frac{5}{2}\right)-k=0$
$\frac{25}{2}-20-k=0$
$k=-\frac{15}{2}$
Sum of roots $=\frac{8}{2}=4$
Other root $=4-\frac{5}{2}=\frac{3}{2}$
Final Answer: $k=-\frac{15}{2}$, other root $=\frac{3}{2}$
Q20 SA-II · 2021
00:00
If the quadratic equation $(1+a^2)x^2+2abx+(b^2-c^2)=0$ has real and equal roots, prove that $b^2=c^2(1+a^2)$.
For equal roots, discriminant $D=0$
$D=(2ab)^2-4(1+a^2)(b^2-c^2)=0$
$4a^2b^2-4(1+a^2)(b^2-c^2)=0$
$a^2b^2=(1+a^2)(b^2-c^2)$
$b^2=c^2(1+a^2)$
Hence proved.
Q21 SA-II · 2020
00:00
Find the nature of roots of $3x^2-4\sqrt{3}x+4=0$. If real, find the roots.
$a=3,\; b=-4\sqrt{3},\; c=4$
$D=b^2-4ac=48-48=0$
Roots are real and equal
Root $=-\frac{b}{2a}=\frac{4\sqrt{3}}{6}=\frac{2\sqrt{3}}{3}$
Final Answer: Roots are equal = $\frac{2\sqrt{3}}{3}$
Q22 SA-II · 2017
00:00
Find the value of $k$ for which $x^2+k(2x+k-1)+2=0$ has real and equal roots.
Simplify equation:
$x^2+2kx+k^2-k+2=0$
$a=1,\; b=2k,\; c=k^2-k+2$
$D=b^2-4ac=4k^2-4(k^2-k+2)$
$4k-8=0 \Rightarrow k=2$
Final Answer: $k=2$
Q23 SA-I · 2019
00:00
Find the value of $k$ for which $3x^2+kx+3=0$ has real and equal roots.
For equal roots: $D=0$
$k^2-36=0$
$k=±6$
Final Answer: $k=±6$
Q24 SA-I · 2025
00:00
Find the value(s) of $k$ so that $4x^2+kx+1=0$ has real and equal roots.
$D=k^2-16=0$
$k=±4$
Final Answer: $k=±4$
Q25 SA-I · 2023
00:00
Find the discriminant of $4x^2-5=0$ and comment on the nature of roots.
$a=4,\; b=0,\; c=-5$
$D=0-4(4)(-5)=80$
Roots are real and distinct.
Q26 SA-I · 2021
00:00
Find the value of $m$ for which $(m-1)x^2+2(m-1)x+1=0$ has real and equal roots.
For equal roots: $D=0$
$[2(m-1)]^2-4(m-1)(1)=0$
$(m-1)(m-2)=0$
$m=1$ or $m=2$
Final Answer: $m=1$ or $m=2$
00:00
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
Sum of roots: $-\frac{b}{a}=\frac{6}{5}$
Product of roots: $\frac{c}{a}=\frac{21}{5}$
Ratio: $\frac{6}{5}:\frac{21}{5}=6:21=2:7$
Final Answer: (b) 2 : 7
Q2 2023
00:00
Which of the following quadratic equations has sum of its roots as 4?
(a)$2x^2-4x+8=0$
(b)$-x^2+4x+4=0$
(c)$\sqrt{2}x^2-\frac{4}{\sqrt{2}}x+1=0$
(d)$4x^2-4x+4=0$
Formula: Sum of roots $=-\frac{b}{a}$
For option (b): $-\frac{4}{-1}=4$
Final Answer: (b)
Q3 2025
00:00
The discriminant of the quadratic equation $bx^2+ax+c=0$ is:
(a)$b^2-4ac$
(b)$\sqrt{b^2-4ac}$
(c)$\sqrt{a^2-4bc}$
(d)$a^2-4bc$
Discriminant: $D=b^2-4ac$
Here coefficient of $x$ is $a$
$D=a^2-4bc$
Final Answer: (d)
Q4 2020
00:00
The discriminant of $3\sqrt{3}x^2+10x+\sqrt{3}=0$ is:
(a)±8
(b)8
(c)$100-4\sqrt{3}$
(d)64
$D=b^2-4ac$
$=100-4(3\sqrt{3})(\sqrt{3})$
$=100-36=64$
Final Answer: (d) 64
Q5 2025
00:00
Which equation has real and equal roots?
(a)$(x+1)^2=2x+1$
(b)$x^2+x=0$
(c)$x^2-4=0$
(d)$x^2+x+1=0$
$(x+1)^2=2x+1 \Rightarrow x^2=0$
Discriminant $=0$
Final Answer: (a)
Q6 2024
00:00
If the roots of $ax^2+bx+c=0$ are real and equal, which relation is true?
(a)$a=\frac{b^2}{c}$
(b)$b^2=ac$
(c)$ac=\frac{b^2}{4}$
(d)$c=\frac{b^2}{a}$
$D=b^2-4ac=0$
$ac=\frac{b^2}{4}$
Final Answer: (c)
Q7 2023
00:00
Values of $k$ for which $4x^2+kx+9=0$ has equal roots:
(a)±11
(b)±12
(c)±6
(d)±3
$D=k^2-144=0$
$k=±12$
Final Answer: (b)
Q8 2023
00:00
Least positive value of $k$ for which $2x^2+kx-4=0$ has rational roots:
(a)±$2\sqrt{2}$
(b)2
(c)±2
(d)$\sqrt{2}$
$D=k^2+32$ must be perfect square
Least such $k=2$
Final Answer: (b)
Q9 2020
00:00
Value(s) of $k$ for which $2x^2+kx+2=0$ has equal roots:
(a)4
(b)±4
(c)−4
(d)0
$D=k^2-16=0$
$k=±4$
Final Answer: (b)
Q10 2019
00:00
Write the discriminant of the quadratic equation $(x+5)^2=2(5x-3)$.
Given: $(x+5)^2=10x-6$
$x^2+10x+25-10x+6=0$
$x^2+31=0$
$D=b^2-4ac=0-124=-124$
Final Answer: Discriminant = −124
Q11 VSA · 2019
00:00
Write the discriminant of the quadratic equation $(x + 5)^2 = 2(5x − 3)$.
$ (x+5)^2 = 10x - 6 $
$ x^2 + 10x + 25 - 10x + 6 = 0 $
$ x^2 + 31 = 0 $
Here, $a=1,\; b=0,\; c=31$
Discriminant $D=b^2-4ac=0-124=-124$
Final Answer: Discriminant = −124
Q12 VSA · 2019
00:00
Find the nature of roots of the quadratic equation $2x^2 − 4x + 3 = 0$.
$a=2,\; b=-4,\; c=3$
$D=b^2-4ac=16-24=-8$
Since $D<0$, the roots are non-real (complex).
Q13 SA-I · 2023
00:00
Find the sum and product of the roots of the quadratic equation $2x^2 − 9x + 4 = 0$.
$a=2,\; b=-9,\; c=4$
Sum of roots $= -\frac{b}{a}=\frac{9}{2}$
Product of roots $=\frac{c}{a}=\frac{4}{2}=2$
Final Answer: Sum = 9/2, Product = 2
Q14 SA-I · 2017
00:00
Find the value of $p$ for which one root of $px^2 − 14x + 8 = 0$ is six times the other.
Let roots be $\alpha$ and $6\alpha$
Sum: $\alpha+6\alpha=7\alpha=\frac{14}{p}$
$\alpha=\frac{2}{p}$
Product: $6\alpha^2=\frac{8}{p}$
$6\left(\frac{4}{p^2}\right)=\frac{8}{p}$
$p=3$
Final Answer: $p=3$
Q15 SA-I · 2019
00:00
If one root of the quadratic equation $2x^2+2x+k=0$ is $-1/3$, find the value of $k$.
Substitute $x=-1/3$
$2\left(\frac{1}{9}\right)-\frac{2}{3}+k=0$
$\frac{2}{9}-\frac{6}{9}+k=0$
$k=\frac{4}{9}$
Final Answer: $k=4/9$
Q16 SA-I · 2019
00:00
Find the value of $k$ for which the roots of $3x^2 − 10x + k = 0$ are reciprocal of each other.
For reciprocal roots: $\frac{c}{a}=1$
$\frac{k}{3}=1$
$k=3$
Final Answer: $k=3$
Q17 SA-I · 2018
00:00
If $x=3$ is one root of $x^2 − 2kx − 6 = 0$, find the value of $k$.
Substitute $x=3$
$9-6k-6=0$
$3=6k$
$k=\frac{1}{2}$
Final Answer: $k=1/2$
Q18 SA-I · 2019
00:00
For what values of $a$ does the quadratic equation $9x^2 − 3ax + 1 = 0$ have equal roots?
Equal roots ⇒ $D=0$
$(-3a)^2-4(9)(1)=0$
$9a^2=36$
$a=±2$
Final Answer: $a=±2$
Q19 SA-II · 2021
00:00
One root of the quadratic equation $2x^2 - 8x - k = 0$ is $\frac{5}{2}$. Find the value of $k$ and the other root.
Given one root $x=\frac{5}{2}$
Substitute in equation:
$2\left(\frac{25}{4}\right)-8\left(\frac{5}{2}\right)-k=0$
$\frac{25}{2}-20-k=0$
$k=-\frac{15}{2}$
Sum of roots $=\frac{8}{2}=4$
Other root $=4-\frac{5}{2}=\frac{3}{2}$
Final Answer: $k=-\frac{15}{2}$, other root $=\frac{3}{2}$
Q20 SA-II · 2021
00:00
If the quadratic equation $(1+a^2)x^2+2abx+(b^2-c^2)=0$ has real and equal roots, prove that $b^2=c^2(1+a^2)$.
For equal roots, discriminant $D=0$
$D=(2ab)^2-4(1+a^2)(b^2-c^2)=0$
$4a^2b^2-4(1+a^2)(b^2-c^2)=0$
$a^2b^2=(1+a^2)(b^2-c^2)$
$b^2=c^2(1+a^2)$
Hence proved.
Q21 SA-II · 2020
00:00
Find the nature of roots of $3x^2-4\sqrt{3}x+4=0$. If real, find the roots.
$a=3,\; b=-4\sqrt{3},\; c=4$
$D=b^2-4ac=48-48=0$
Roots are real and equal
Root $=-\frac{b}{2a}=\frac{4\sqrt{3}}{6}=\frac{2\sqrt{3}}{3}$
Final Answer: Roots are equal = $\frac{2\sqrt{3}}{3}$
Q22 SA-II · 2017
00:00
Find the value of $k$ for which $x^2+k(2x+k-1)+2=0$ has real and equal roots.
Simplify equation:
$x^2+2kx+k^2-k+2=0$
$a=1,\; b=2k,\; c=k^2-k+2$
$D=b^2-4ac=4k^2-4(k^2-k+2)$
$4k-8=0 \Rightarrow k=2$
Final Answer: $k=2$
Q23 SA-I · 2019
00:00
Find the value of $k$ for which $3x^2+kx+3=0$ has real and equal roots.
For equal roots: $D=0$
$k^2-36=0$
$k=±6$
Final Answer: $k=±6$
Q24 SA-I · 2025
00:00
Find the value(s) of $k$ so that $4x^2+kx+1=0$ has real and equal roots.
$D=k^2-16=0$
$k=±4$
Final Answer: $k=±4$
Q25 SA-I · 2023
00:00
Find the discriminant of $4x^2-5=0$ and comment on the nature of roots.
$a=4,\; b=0,\; c=-5$
$D=0-4(4)(-5)=80$
Roots are real and distinct.
Q26 SA-I · 2021
00:00
Find the value of $m$ for which $(m-1)x^2+2(m-1)x+1=0$ has real and equal roots.
For equal roots: $D=0$
$[2(m-1)]^2-4(m-1)(1)=0$
$(m-1)(m-2)=0$
$m=1$ or $m=2$
Final Answer: $m=1$ or $m=2$
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