JEE Main 2024MathematicsQuadratic EquationHardNumerical

JEE Main 2024Quadratic Equation Question with Solution

JEE Main 2024 (06 Apr Shift 1)

Question

Let be the solution of the equation and . Then the value of is

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Show full solutionCorrect answer: 221
Correct answer
221

Step-by-step explanation

$\begin{aligned} & 4 x^4+8 x^3-17 x^2-12 x+9 \\ & =4\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right) \end{aligned}x=2 i \&-2 i$ $\begin{aligned} & 64-64 i+68-24 i+9=\left(2 i-x_1\right)\left(2 i-x_2\right)\left(2 i-x_3\right) \\ & \left(2 i-x_4\right) \\ & =141-88 i ....(1)\\ & 64+64 i+68+24 i+9=4\left(-2 i-x_1\right)\left(-2 i-x_2\right)(-2 i \\ & \left.-x_3\right)\left(-2 i-x_4\right) \\ & =141+88 i....(2) \end{aligned}$ $\begin{aligned} & \frac{125}{16} \mathrm{~m}=\frac{141^2+88^2}{16} \\ & \mathrm{~m}=221 \end{aligned}$

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About this question

This is a previous-year question from JEE Main 2024, covering the Quadratic Equation chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.