JEE Main 2024 — Quadratic 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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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.