Mathematics · Quadratic Equations

JEE Main 2024 — 6 April, Shift 1 — Question 24

Let x1,x2,x3,x4x_{1}, x_{2}, x_{3}, x_{4} be the solution of the equation 4x4+8x3−17x2−12x+9=04 x^{4}+8 x^{3}-17 x^{2}-12 x+9=0 and (4+x12)(4+x22)(4+x32)(4+x42)=12516m\left(4+x_{1}^{2}\right)\left(4+x_{2}^{2}\right)\left(4+x_{3}^{2}\right)\left(4+x_{4}^{2}\right)=\frac{125}{16} m . Then the value of mm is \qquad

Answer: 221

Numerical answer — enter this value.

Step-by-step solution

4x4+8x3−17x2−12x+94 \mathrm{x}^{4}+8 \mathrm{x}^{3}-17 \mathrm{x}^{2}-12 \mathrm{x}+9

=4(x−x1)(x−x2)(x−x3)(x−x4)=4\left(\mathrm{x}-\mathrm{x}_{1}\right)\left(\mathrm{x}-\mathrm{x}_{2}\right)\left(\mathrm{x}-\mathrm{x}_{3}\right)\left(\mathrm{x}-\mathrm{x}_{4}\right)

Put x=2i&−2ix=2 \mathrm{i} \&-2\mathrm{i}

64−64i+68−24i+9=(2i−x1)(2i−x2)(2i−x3)64-64 \mathrm{i}+68-24 \mathrm{i}+9=\left(2 \mathrm{i}-\mathrm{x}_{1}\right)\left(2 \mathrm{i}-\mathrm{x}_{2}\right)\left(2 \mathrm{i}-\mathrm{x}_{3}\right)

(2i−x4)\left(2 \mathrm{i}-\mathrm{x}_{4}\right) =141−88i=141-88 \mathrm{i}

64+64i+68+24i+9=4(−2i−x1)(−2i−x2)(−2i−x3)64+64 \mathrm{i}+68+24 \mathrm{i}+9=4\left(-2 \mathrm{i}-\mathrm{x}_{1}\right)\left(-2 \mathrm{i}-\mathrm{x}_{2}\right)(-2\mathrm{i} \left.-\mathrm{x}_{3}\right)

(−2i−x4)=141+88i\left(-2 \mathrm{i}-\mathrm{x}_{4}\right) =141+88 i

12516 m=1412+88216\frac{125}{16} \mathrm{~m}=\frac{141^{2}+88^{2}}{16}

m=221m=221

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Quadratic Equations
Topic
Theory of Equations of Degree other than 2
Let x 1 , x 2 , x 3 , x 4 be the solution of the equation 4 x 4 +8 x… | JEE Main 2024 PYQ with Solution · DhiX AI