Quadratic Equation JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Quadratic Equation, free to read — no sign-in needed. The full chapter has 19 questions; sign in to attempt the remaining 14 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Let $a, b, c$ be positive integers in arithmetic progression such that the equation $ax^2 + bx + c = 0$ has only integer solutions. Then which of the following statements is (are) TRUE ?
   1. A.$c - b$ is an integer multiple of $a$
   2. B.Both the roots of the equation $ax^2 + bx + c = 0$ are odd integers
   3. C.If $c = 15$, then $ab = 8$
   4. D.If $b = 8$, then $x = 3$ is a root of the equation $ax^2 + bx + c = 0$
   Show answer & solution

   Answer: A,B,C

   Given $a, b, c$ are in arithmetic progression, we have $2b = a + c$. Let the integer roots of the equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$. Sum of the roots is $\alpha + \beta = -\dfrac{b}{a}$ and product of the roots is $\alpha \beta = \dfrac{c}{a}$. Dividing the arithmetic progression condition by $a$, we get: $\dfrac{2b}{a} = 1 + \dfrac{c}{a}$ Substituting the sum and product of the roots: $-2(\alpha + \beta) = 1 + \alpha \beta$ $\alpha \beta + 2\alpha + 2\beta + 1 = 0$ Adding $3$ to both sides to factorize: $\alpha \beta + 2\alpha + 2\beta + 4 = 3$ $(\alpha + 2)(\beta + 2) = 3$ Since $\alpha$ and $\beta$ are integers, $(\alpha + 2)$ and $(\beta + 2)$ must be integer factors of $3$. The possible pairs are $(1, 3)$, $(3, 1)$, $(-1, -3)$, and $(-3, -1)$. If $(\alpha + 2, \beta + 2) = (1, 3)$ or $(3, 1)$, the roots are $-1$ and $1$. The sum of the roots is $0$, which implies $b = 0$. This is rejected because $b$ is a positive integer. If $(\alpha + 2, \beta + 2) = (-1, -3)$ or $(-3, -1)$, the roots are $-3$ and $-5$. The sum of the roots is $-8$, which implies $-\dfrac{b}{a} = -8 \Rightarrow b = 8a$. The product of the roots is $15$, which implies $\dfrac{c}{a} = 15 \Rightarrow c = 15a$. Since $a$ is a positive integer, $b = 8a$ and $c = 15a$ are also positive integers. The roots of the equation are always $-3$ and $-5$. Evaluating the given statements: $c - b = 15a - 8a = 7a$, which is an integer multiple of $a$. The roots are $-3$ and $-5$, both of which are odd integers. If $c = 15$, then $15a = 15 \Rightarrow a = 1$. Consequently, $b = 8(1) = 8$. Thus, $ab = 1 \times 8 = 8$. If $b = 8$, then $8a = 8 \Rightarrow a = 1$. The roots are $-3$ and $-5$, so $x = 3$ is not a root. Answer: $c - b$ is an integer multiple of $a$; Both the roots of the equation $ax^2 + bx + c = 0$ are odd integers; If $c = 15$, then $ab = 8$
2. Q2JEE Advanced Adv 2024 (Paper 1)
   Let $f(x)=x^4+a x^3+b x^2+c$ be a polynomial with real coefficients such that $f(1)=-9$. Suppose that $i \sqrt{3}$ is a root of the equation $4 x^3+3 a x^2+2 b x=0$, where $i=\sqrt{-1}$. If $\alpha_1, \alpha_2, \alpha_3$, and $\alpha_4$ are all the roots of the equation $f(x)=0$, then $\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2$ is equal to ________.
   Show answer & solution

   Answer: 20

   $f(1)=1+a+b+c=-9 \quad \Rightarrow \quad a+b+c=-10.....(1)$ $4 x^3+3 a x^2+2 b x=0$ roots are $\sqrt{3} i,-\sqrt{3} i, 0$ $\Rightarrow4 x^2+3 a x+2 b=0 \lt ^{\sqrt{3i}}_{-{\sqrt{3i}}}$ $\Rightarrow \quad \mathrm{a}=0 ~\&~ \frac{2 \mathrm{~b}}{4}=(\sqrt{3} \mathrm{i})(-\sqrt{3} \mathrm{i})$ $\mathrm{b}=6$ use $\mathrm{a}, \mathrm{b}$ in (1) $\Rightarrow \mathrm{c}=-16$ $\begin{array}{ll}\Rightarrow & \mathrm{f}(\mathrm{x})=\mathrm{x}^4+6 \mathrm{x}^2-16=0 \\ & \left(\mathrm{x}^2+8\right)\left(\mathrm{x}^2-2\right)=0 \\ \Rightarrow \quad & \mathrm{x}= \pm \sqrt{8} \mathrm{i}, \pm \sqrt{2}\end{array}$ $\Rightarrow \quad\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2=20$
3. Q3JEE Advanced Adv 2021 (Paper 1)
   For $x\in R$, the number of real roots of the equation $3{x}^{2}-4\left|{x}^{2}-1\right|+x-1=0$ is
   Show answer & solution

   Answer: 4

   $\left|{x}^{2}-1\right|=\left\{\begin{matrix}{x}^{2}-1,{x}^{2}\geq 1 \\ -\left({x}^{2}-1\right){x}^{2}\lt 1\end{matrix}\right.$ Case 1 ${x}^{2}-1\lt 0\Rightarrow x\in (-1,1)$ $3{x}^{2}+4\left({x}^{2}-1\right)+x-1=0$ $7{x}^{2}+x-5=0$ $x=\dfrac{-1\pm \sqrt{1+140}}{2\times 7}=\dfrac{-1+\sqrt{141}}{14},\dfrac{-1-\sqrt{141}}{14}$ We know, $\sqrt{141}\approx 11.5$ So, both roots lie in the interval $\left(-1,1\right)$ Case 2 ${x}^{2}-1\geq 0\Rightarrow x\in (-\infty ,-1]\cup [1,\infty )$ $3{x}^{2}-4\left({x}^{2}-1\right)+x-1=0$ $-{x}^{2}+4+x-1=0$ ${x}^{2}-x-3=0$ $x=\dfrac{1\pm \sqrt{1+12}}{2}=\dfrac{1+\sqrt{13}}{2},\dfrac{1-\sqrt{13}}{2}$ We know, $3\lt \sqrt{13}\lt 4$ So, both roots lie in the interval $(-\infty ,-1]\cup [1,\infty )$ Hence, $4$ real roots.
4. Q4JEE Advanced Adv 2020 (Paper 1)
   Suppose $a,b$ denote the distinct real roots of the quadratic polynomial ${x}^{2}+20x-2020$ and suppose $c,d$ denote the distinct complex roots of the quadratic polynomial ${x}^{2}-20x+2020$. Then the value of $ac\left(a-c\right)+ad\left(a-d\right)+bc\left(b-c\right)+bd\left(b-d\right)$
   1. A.$0$
   2. B.$8000$
   3. C.$8080$
   4. D.$16000$
   Show answer & solution

   Answer: (D)

   Finding the sum and product of roots of the two given quadratic equations i.e. $a+b=-20\&ab=-2020...........\left(i\right)$ $c+d=20\&cd=2020...........\left(ii\right)$ Now, $ac\left(a-c\right)+ad\left(a-d\right)+bc\left(b-c\right)+bd\left(b-d\right)$ $={a}^{2}\left(c+d\right)-a\left({c}^{2}+{d}^{2}\right)+{b}^{2}\left(c+d\right)-b\left({c}^{2}+{d}^{2}\right)$ $=\left({a}^{2}+{b}^{2}\right)\left(c+d\right)-\left(a+b\right)\left({c}^{2}+{d}^{2}\right)$ $=\left\{{\left(a+b\right)}^{2}-2ab\right\}\left(c+d\right)-\left(a+b\right)\left\{{\left(c+d\right)}^{2}-2cd\right\}$ $=\left\{{\left(-20\right)}^{2}-2\left(-2020\right)\right\}\left(20\right)-\left(-20\right)\left\{{\left(20\right)}^{2}-2\left(2020\right)\right\}$ (using equations $\left(i\right)\&\left(ii\right)$) $=16000$
5. Q5JEE Advanced Adv 2017 (Paper 2)
   Paragraph: Let $p, q$ be integers and let $\alpha, \beta$ be the roots of the equation, $x^{2}-x-1=0$, where $\alpha \neq \beta$. For $n=0,1,2, \ldots$, let $a_{n}=p \alpha^{n}+q \beta^{n}$ FACT: If $a$ and $b$ are rational numbers and $a+b \sqrt{5}=0$, then $a=0=b$. Question: $a_{12}=$
   1. A.$2{a}_{11}+{a}_{10}$
   2. B.${a}_{11}-{a}_{10}$
   3. C.${a}_{11}+{a}_{10}$
   4. D.${a}_{11}+2{a}_{10}$
   Show answer & solution

   Answer: (C)

   ${\alpha }^{2}=\alpha +1\Rightarrow {\alpha }^{n}={\alpha }^{n-1}+{\alpha }^{n-2}$ $\Rightarrow p{\alpha }^{n}+q{\beta }^{n}=p\left({\alpha }^{n-1}+{\alpha }^{n-2}\right)+q\left({\beta }^{n-1}+{\beta }^{n-2}\right)$ ${a}_{n}={a}_{n-1}+{a}_{n-2}$ $\Rightarrow {a}_{12}={a}_{11}+{a}_{10}$