JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 21)
Let $$x_1, x_2, x_3, x_4$$ be the solution of the equation $$4 x^4+8 x^3-17 x^2-12 x+9=0$$ and $$\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$$. Then the value of $$m$$ is _________.
Answer
221
Explanation
$$\begin{aligned} & 4 x^4+8 x^3-17 x^2-12 x+9=0 \\ & (x+1)\left(4 x^3+4 x^2-21 x+9\right)=0 \\ & (x+1)(x+3)\left(4 x^2-8 x+3\right)=0 \\ & (x+1)(x+3)\left(4 x^2-6 x-2 x+3\right)=0 \\ & (x+1)(x+3)(2 x(2 x-3)-1(2 x-3))=0 \end{aligned}$$
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$$\begin{aligned} & (4+1)(4+9)\left(4+\frac{1}{4}\right)\left(4+\frac{9}{4}\right)=\frac{125}{16} m \\ & 5 \times 13 \times\left(\frac{17}{4}\right) \times\left(\frac{25}{4}\right)=\frac{125}{16} m \\ & \frac{125}{16} \times[13 \times 17]=\frac{125}{16} m \\ & m=13 \times 17 \\ & m=221 \end{aligned}$$
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