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QJEEM24Maecfb: JEE Main 2024 Mathematics - Class 11 JEEM24Maecfb Solution: Let be the solution of the equation and . Then the value of is…

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 _________.

Step-by-Step Verified Solution

amp;4x4+8x3−17x2−12x+9=0amp;(x+1)(4x3+4x2−21x+9)=0amp;(x+1)(x+3)(4x2−8x+3)=0amp;(x+1)(x+3)(4x2−6x−2x+3)=0amp;(x+1)(x+3)(2x(2x−3)−1(2x−3))=0\begin{aligned}& 4 x^4+8 x^3-17 x^2-12 x+9=0 \\& (x+1)\left(4 x^3+4 x^2-21 x+9\right)=0 \\& (x+1)(x+3)\left(4 x^2-8 x+3\right)=0 \\& (x+1)(x+3)\left(4 x^2-6 x-2 x+3\right)=0 \\& (x+1)(x+3)(2 x(2 x-3)-1(2 x-3))=0\end{aligned}

JEE Main 2024 (Online) 6th April Morning Shift Mathematics - Quadratic Equation and Inequalities Question 9 English Explanation

amp;(4+1)(4+9)(4+14)(4+94)=12516mamp;5×13×(174)×(254)=12516mamp;12516×[13×17]=12516mamp;m=13×17amp;m=221\begin{aligned}& (4+1)(4+9)\left(4+\frac{1}{4}\right)\left(4+\frac{9}{4}\right)=\frac{125}{16} m \\& 5 \times 13 \times\left(\frac{17}{4}\right) \times\left(\frac{25}{4}\right)=\frac{125}{16} m \\& \frac{125}{16} \times[13 \times 17]=\frac{125}{16} m \\& m=13 \times 17 \\& m=221\end{aligned}

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