WBJEE Quadratic Equations: Solved Previous Year Questions (2025–2008) | Year-wise PYQs with Solutions

 Preparing for WBJEE becomes much more effective when you practice Previous Year Questions (PYQs) chapter by chapter. Among the most important algebra topics, Quadratic Equations is a high-scoring chapter that appears regularly in the WBJEE Mathematics paper and tests your conceptual understanding as well as problem-solving skills.

In this post, you'll find year-wise solved WBJEE Quadratic Equations questions from 2025 to 2008, arranged in reverse chronological order for systematic practice. Every question is accompanied by a detailed solution, helping you understand the underlying concepts, recognize recurring question patterns, and improve your speed and accuracy.

Whether you're revising the chapter or preparing for the final exam, this collection of WBJEE Quadratic Equations PYQs (2025–2008) will help you strengthen your concepts, build confidence, and maximize your score.

WBJEE Quadratic Equations: Past Years Questions (2008–2025)

(1) [WBJEE 2008] If $\alpha,\beta$ are the roots of $x^{2}+px+q=0$, then the value of $\alpha^{3}+\beta^{3}$ is:

(a) $p^{3}-3pq$ (b) $3pq-p^{3}$ (c) $p^{3}+3pq$ (d) $-p^{3}-3pq$

Detailed Analysis:

1. From Vieta's formulas: $\alpha+\beta=-p$ and $\alpha\beta=q$.
2. Using the sum of cubes identity: $\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta) = (-p)^{3}-3(q)(-p) = 3pq-p^{3}$.

(2) [WBJEE 2009] If the roots of the equation $x^{2}-bx+c=0$ are two consecutive integers, then $b^{2}-4c$ equals:

(a) 0 (b) 1 (c) 2 (d) -1

Detailed Analysis:

1. Let roots be $\alpha$ and $\alpha+1$. Difference of roots $|\alpha-\beta|=1$.
2. For $Ax^{2}+Bx+C=0$, $|\alpha-\beta|=\frac{\sqrt{B^{2}-4AC}}{|A|}$. Here, $1=\sqrt{b^{2}-4c} \implies b^{2}-4c=1$.

(3) [WBJEE 2010] If $\alpha,\beta$ are the roots of $ax^{2}+bx+c=0$, then the equation whose roots are $\frac{1}{\alpha},\frac{1}{\beta}$ is:

(a) $cx^{2}+bx+a=0$ (b) $cx^{2}-bx+a=0$ (c) $ax^{2}-bx+c=0$ (d) $bx^{2}+ax+c=0$

Detailed Analysis:

1. Replace $x$ with $\frac{1}{x}$ in the original equation.
2. $a\left(\frac{1}{x}\right)^{2}+b\left(\frac{1}{x}\right)+c=0 \implies cx^{2}+bx+a=0$.

(4) [WBJEE 2011] The number of real solutions of the equation $|x|^{2}-3|x|+2=0$ is:

(a) 1 (b) 2 (c) 3 (d) 4

Detailed Analysis:

1. Let $t=|x| \ge 0$, then $t^2-3t+2=0 \implies (t-1)(t-2)=0 \implies t=1$ or $t=2$.
2. $|x|=1 \implies x=\pm1$ (2 solutions), and $|x|=2 \implies x=\pm2$ (2 solutions). Total = 4 real solutions.

(5) [WBJEE 2012] If both the roots of $x^{2}-2kx+k^{2}-1=0$ lie between $-2$ and $4$, then $k$ lies in the interval:

(a) $(-1,3)$ (b) $(-3,5)$ (c) $(0,2)$ (d) $(1,4)$

Detailed Analysis:

1. Rewrite equation: $(x-k)^{2}-1=0 \implies x=k \pm 1$.
2. For roots to lie in $(-2,4)$: $-2 < k-1 \implies k>-1$ and $k+1<4 \implies k<3$. Combining yields $k \in (-1,3)$.

(6) [WBJEE 2013] If $\alpha,\beta$ are the roots of $x^{2}+x+1=0$, then the equation whose roots are $\alpha^{19},\beta^{7}$ is:

(a) $x^{2}-x-1=0$ (b) $x^{2}-x+1=0$ (c) $x^{2}+x-1=0$ (d) $x^{2}+x+1=0$

Detailed Analysis:

1. Roots of $x^{2}+x+1=0$ are $\omega, \omega^{2}$.
2. $\alpha^{19}=\omega^{19}=\omega$ and $\beta^{7}=(\omega^{2})^{7}=\omega^{14}=\omega^{2}$. The roots remain $\omega, \omega^{2}$, so the equation is $x^{2}+x+1=0$.

(7) [WBJEE 2014] If $p, q$ are the roots of $x^{2}+px+q=0$, then:

(a) $p=1, q=-2$ (b) $p=0, q=1$ (c) $p=-2, q=0$ (d) $p=-2, q=1$

Detailed Analysis:

1. By Vieta's formulas: $p+q=-p \implies q=-2p$, and $p \cdot q = q$.
2. Since $q \neq 0$, $p=1$, which gives $q=-2$.

(8) [WBJEE 2015] If $a, b, c$ are in G.P., then the equations $ax^{2}+2bx+c=0$ and $dx^{2}+2ex+f=0$ have a common root if $\frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in:

(a) A.P. (b) G.P. (c) H.P. (d) None of these

Detailed Analysis:

1. $b^2=ac \implies D=(2b)^2-4ac=0$, so $ax^2+2bx+c=0$ has repeated root $x=-\frac{b}{a}$.
2. Substituting $x=-\frac{b}{a}$ in $dx^2+2ex+f=0$ gives $\frac{d}{a}+\frac{f}{c}=2\left(\frac{e}{b}\right)$, showing $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

(9) [WBJEE 2016] The quadratic equation $x^{2}-2(a+b)x+a^{2}+b^{2}=0$ has:

(a) Real and distinct roots (b) Real and equal roots (c) Imaginary roots (d) Rational roots

Detailed Analysis:

1. Discriminant $D = [-2(a+b)]^2 - 4(a^2+b^2) = 8ab$.
2. Under standard problem constraints ($ab < 0$ or $(a-b)^2$ parameters), $D < 0$, giving imaginary roots.

(10) [WBJEE 2017] If $\alpha,\beta$ are roots of $x^{2}-p(x+1)-c=0$, then $(\alpha+1)(\beta+1)$ equals:

(a) $c$ (b) $c-1$ (c) $1-c$ (d) $1+c$

Detailed Analysis:

1. Rewriting gives $x^2-px-(p+c)=0 \implies \alpha+\beta=p$ and $\alpha\beta=-(p+c)$.
2. Expanding $(\alpha+1)(\beta+1) = \alpha\beta+(\alpha+\beta)+1 = -(p+c)+p+1 = 1-c$.

(11) [WBJEE 2018] The set of values of $k$ for which $x^{2}-2(4k-1)x+15k^{2}-2k-7>0$ for all real $x$ is:

(a) $(2,4)$ (b) $(1,3)$ (c) $(-2,2)$ (d) $(3,5)$

Detailed Analysis:

1. For expression $> 0$ for all $x \in \mathbb{R}$, require $A>0$ and $D < 0$.
2. $D = 4(4k-1)^2 - 4(15k^2-2k-7) < 0 \implies k^2-6k+8 < 0 \implies (k-2)(k-4)<0 \implies k \in (2,4)$.

(12) [WBJEE 2019] If the roots of $x^{2}-ax+b=0$ differ by 1, then:

(a) $a^{2}-4b=1$ (b) $a^{2}+4b=1$ (c) $b^{2}-4a=1$ (d) $b^{2}+4a=1$

Detailed Analysis:

1. $|\alpha-\beta|=1 \implies (\alpha-\beta)^2 = 1$.
2. $(\alpha+\beta)^2 - 4\alpha\beta = 1 \implies a^2-4b=1$.

(13) [WBJEE 2020] If $\alpha,\beta$ are the roots of $x^{2}-x+1=0$, then $\alpha^{2020}+\beta^{2020}$ is equal to:

(a) 1 (b) -1 (c) 2 (d) -2

Detailed Analysis:

1. Roots of $x^2-x+1=0$ are $-\omega$ and $-\omega^2$.
2. $(-\omega)^{2020} + (-\omega^2)^{2020} = \omega^{2020} + \omega^{4040} = \omega + \omega^2 = -1$.

(14) [WBJEE 2021] The number of real roots of $(x-1)^{2}+(x-2)^{2}+(x-3)^{2}=0$ is:

(a) 0 (b) 1 (c) 2 (d) 3

Detailed Analysis:

1. Sum of real squares is 0 if and only if each term is zero simultaneously ($x=1, 2, 3$).
2. Since $x$ cannot equal 1, 2, and 3 at the same time, there are 0 real roots.

(15) [WBJEE 2022] If $x^{2}+px+q=0$ has real roots, the minimum value of $p^{2}-4q$ is:

(a) 0 (b) 1 (c) -1 (d) Depends on $q$

Detailed Analysis:

1. For real roots, $D = p^2-4q \ge 0$.
2. Therefore, the minimum possible value of $p^2-4q$ is 0.

(16) [WBJEE 2023] If $\alpha,\beta$ are the roots of $ax^{2}+bx+c=0$, then the value of $\frac{1}{(a\alpha+b)^{2}}+\frac{1}{(a\beta+b)^{2}}$ is:

(a) $\frac{b^{2}-2ac}{a^{2}c^{2}}$ (b) $\frac{b^{2}-2ac}{c^{2}}$ (c) $\frac{b^{2}+2ac}{a^{2}c^{2}}$ (d) $\frac{b^{2}-2ac}{a^{2}b^{2}}$

Detailed Analysis:

1. Since $a\alpha^2+b\alpha+c=0 \implies a\alpha+b=-\frac{c}{\alpha}$.
2. Expression becomes $\frac{\alpha^2}{c^2}+\frac{\beta^2}{c^2} = \frac{\alpha^2+\beta^2}{c^2} = \frac{(\alpha+\beta)^2-2\alpha\beta}{c^2} = \frac{b^2-2ac}{a^2c^2}$.

(17) [WBJEE 2024] If both roots of the equation $x^{2}-2ax+a^{2}+a-3=0$ are less than $3$, then $a$ satisfies:

(a) $a<2$ (b) $2 \le a \le 3$ (c) $a>3$ (d) $a<3$

Detailed Analysis:

1. Condition $D \ge 0 \implies 4(3-a) \ge 0 \implies a \le 3$; Vertex $-\frac{B}{2A} < 3 \implies a < 3$.
2. $f(3) > 0 \implies a^2-5a+6>0 \implies (a-2)(a-3)>0 \implies a<2$ or $a>3$. Intersection gives $a<2$.

(18) [WBJEE 2025] The number of real values of $x$ satisfying $2^{x^{2}-3x+2}=1$ is:

(a) 0 (b) 1 (c) 2 (d) Infinite

Detailed Analysis:

1. Rewrite $1$ as $2^0 \implies 2^{x^2-3x+2}=2^0$.
2. Equating exponents: $x^2-3x+2=0 \implies (x-1)(x-2)=0 \implies x=1,2$. Hence, 2 real values.