Quadratic Equation JEE Main previous year questions with solutions

5 solved JEE Main questions on Quadratic Equation, free to read — no sign-in needed. The full chapter has 199 questions; sign in to attempt the remaining 194 in the exam simulator.

1. Q1JEE Main 2026 (02 Apr, Shift 1)Graph and Sign of Quadratic
   Let $a, b, c \in \{1, 2, 3, 4\}$. If the probability, that $ax^2 + 2\sqrt{2}\,bx + c \gt 0$ for all $x \in \mathbb{R}$, is $\dfrac{m}{n}$, $\gcd(m, n) = 1$, then $m + n$ is equal to _______.
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   Answer: 81

   For the quadratic expression $ax^2 + 2\sqrt{2}bx + c \gt 0$ for all $x \in \mathbb{R}$, the leading coefficient must be positive and the discriminant must be negative. Since $a \in \{1, 2, 3, 4\}$, the condition $a \gt 0$ is always satisfied. The discriminant condition is: $D \lt 0$ $\Rightarrow (2\sqrt{2}b)^2 - 4ac \lt 0$ $\Rightarrow 8b^2 - 4ac \lt 0$ $\Rightarrow 2b^2 \lt ac$ The total number of possible triplets $(a, b, c)$ is $4 \times 4 \times 4 = 64$. Now, we find the number of favorable outcomes by checking the possible values of $b \in \{1, 2, 3, 4\}$: Case 1: $b = 1$ We need $2(1)^2 \lt ac \Rightarrow ac \gt 2$. The total number of pairs $(a, c)$ is $4 \times 4 = 16$. The pairs for which $ac \le 2$ are $(1, 1), (1, 2), (2, 1)$, which are $3$ in number. So, the number of pairs with $ac \gt 2$ is $16 - 3 = 13$. Case 2: $b = 2$ We need $2(2)^2 \lt ac \Rightarrow ac \gt 8$. The possible pairs $(a, c)$ from the given set are $(3, 3), (3, 4), (4, 3), (4, 4)$. So, there are $4$ pairs. Case 3: $b = 3$ We need $2(3)^2 \lt ac \Rightarrow ac \gt 18$. Since the maximum possible value of $ac$ is $4 \times 4 = 16$, there are $0$ pairs. Case 4: $b = 4$ We need $2(4)^2 \lt ac \Rightarrow ac \gt 32$. Again, there are $0$ pairs. Total number of favorable outcomes = $13 + 4 = 17$. The required probability is $\dfrac{m}{n} = \dfrac{17}{64}$. Since $\gcd(17, 64) = 1$, we have $m = 17$ and $n = 64$. Therefore, $m + n = 17 + 64 = 81$. Answer: $81$
2. Q2JEE Main 2025 (23 Jan, Shift 1)Common Roots
   If the equation $\mathrm{a}(\mathrm{b}-\mathrm{c}) \mathrm{x}^2+\mathrm{b}(\mathrm{c}-\mathrm{a}) \mathrm{x}+\mathrm{c}(\mathrm{a}-\mathrm{b})=0$ has equal roots, where $\mathrm{a}+\mathrm{c}=15$ and $\mathrm{b}=\frac{36}{5}$, then $a^2+c^2$ is equal to
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   Answer: 117

   $\begin{aligned} & a(b-c) x^2+b(c-a) x+c(a-b)=0 \\ & x=1 \text { is root } \therefore \text { other root is } 1 \\ & \alpha+\beta=-\frac{b(c-a)}{a(b-c)}=2 \\ & \Rightarrow-\mathrm{bc}+\mathrm{ab}=2 \mathrm{ab}-2 \mathrm{ac} \\ & \Rightarrow 2 \mathrm{ac}=\mathrm{ab}+\mathrm{bc} \\ & \Rightarrow 2 \mathrm{ac}=\mathrm{b}(\mathrm{a}+\mathrm{c}) \\ & \Rightarrow 2 \mathrm{ac}=15 \mathrm{~b} \ldots(1) \\ & \Rightarrow 2 \mathrm{ac}=15\left(\frac{36}{5}\right)=108 \\ & \Rightarrow \mathrm{ac}=54 \\ & \mathrm{a}+\mathrm{c}=15 \\ & \mathrm{a}^2+\mathrm{c}^2+2 \mathrm{ac}=225 \\ & \mathrm{a}^2+\mathrm{c}^2=225-108=117\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 2)Location of Roots
   <p>The number of distinct real roots of the equation $|x+1||x+3|-4|x+2|+5=0$, is</p>
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   Answer: 2

   <p>$|x+1||x+3|-4|x+2|+5=0$ case-1 $\begin{aligned} & x \leq-3 \\ & (x+1)(x+3)+4(x+2)+5=0 \\ & x^2+4 x+3+4 x+8+5=0 \\ & x^2+8 x+16=0 \\ & (x+4)^2=0 \\ & x=-4 \end{aligned}$ case-2 $\begin{aligned} & -3 \leq x \leq-2 \\ & -x^2-4 x-3+4 x+8+5=0 \\ & -x^2+10=0 \\ & x= \pm \sqrt{10} \end{aligned}$ case-3 $\begin{aligned} & -2 \leq x \leq-1 \\ & -x^2-4 x-3-4 x-8+5=0 \\ & -x^2-8 x-6=0 \\ & x^2+8 x+6=0 \\ & x=\frac{-8 \pm 2 \sqrt{10}}{2}=-4 \pm \sqrt{10} \end{aligned}$ case-4 $\begin{aligned} & x \geq-1 \\ & x^2+4 x+3-4 x-8+5=0 \\ & x^2=0 \\ & x=0 \end{aligned}$ No. of solution $=2$</p>
4. Q4JEE Main 2023 (11 Apr, Shift 2)N degree equation
   The number of points, where the curve $f(x)={e}^{8x}-{e}^{6x}-3{e}^{4x}-{e}^{2x}+1,x\in ℝ$ cuts $x$-axis, is equal to............
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   Answer: 2

   Given, ${e}^{8x}-{e}^{6x}-3{e}^{4x}-{e}^{2x}+1=0$ $\Rightarrow {e}^{4x}-{e}^{2x}-3-\dfrac{1}{{e}^{2x}}+\dfrac{1}{{e}^{4x}}=0$ $\Rightarrow {e}^{4x}+\dfrac{1}{{e}^{4x}}+2-\left({e}^{2x}+\dfrac{1}{{e}^{2x}}\right)=5$ $\Rightarrow {\left({e}^{2x}+\dfrac{1}{{e}^{2x}}\right)}^{2}-\left({e}^{2x}+\dfrac{1}{{e}^{2x}}\right)=3$ Now let $\left({e}^{2x}+\dfrac{1}{{e}^{2x}}\right)=t$ So, the equation becomes ${t}^{2}-t-5=0$ $\Rightarrow t=\dfrac{1+\sqrt{21}}{2}$, ignoring negative sign as exponential function are positive, Now ${e}^{2x}+\dfrac{1}{{e}^{2x}}=\dfrac{1+\sqrt{21}}{2}$ $\Rightarrow {e}^{4x}-\dfrac{1+\sqrt{21}}{2}{e}^{2x}+1=0$ which is quadratic equation in ${e}^{2x}$ with upward parabola, Now by $A.M\geq G.M$ we get, ${e}^{2x}+\dfrac{1}{{e}^{2x}}\geq 2$ So, $y=\dfrac{1+\sqrt{21}}{2}\gt 2$, hence it will cut at two distinct point, Hence, there will be two solution.
5. Q5JEE Main 2022 (26 Jul, Shift 2)Range of Quadratic Function
   The minimum value of the sum of the squares of the roots of ${x}^{2}+\left(3-a\right)x=2a-1$ is
   1. A.$6$
   2. B.$4$
   3. C.$5$
   4. D.$8$
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   Answer: (A)

   Let $\alpha ,\beta$ be the roots of the equation ${x}^{2}+\left(3-a\right)x+1-2a=0$ Then, sum of roots $\alpha +\beta =a-3$ And product of roots $\alpha \beta =1-2a$ We know that ${\alpha }^{2}+{\beta }^{2}={\left(\alpha +\beta \right)}^{2}-2\alpha \beta$ $∴{\alpha }^{2}+{\beta }^{2}={\left(a-3\right)}^{2}-2\left(1-2a\right)$ $={a}^{2}-2a+7$ $={\left(a-1\right)}^{2}+6$ $∴$ Minimum value of ${\alpha }^{2}+{\beta }^{2}=6$ at $a=1$.