JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
Let $x_1,x_2,x_3,x_4$ be the solution of the equation $4x^{4}+8x^{3}-17x^{2}-12x+9=0$ 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 $\_\_\_\_\_\_$ .
Quadratic Equation · Higher Degree Polynomial Equations
Answer$221$
Solution
Concept. $4+x_i^2=(x_i-2i)(x_i+2i)$, so the product is the polynomial evaluated at $\pm2i$. Formulas used. (1) $P(x)=4\prod_{k=1}^{4}(x-x_k)$(2) $(2i-x)(-2i-x)=x^2+4$Step 1 - Rewrite. By (1) and (2), $$P(2i)\,P(-2i)=16\prod_{k=1}^{4}(2i-x_k)(-2i-x_k)=16\prod_{k=1}^{4}\left(4+x_k^2\right).$$ Step 2 - Evaluate $P(2i)$. $$P(2i)=64-64i+68-24i+9=141-88i.$$ $P(-2i)$ is its conjugate, so $P(2i)P(-2i)=141^2+88^2=27625$. Step 3 - Find $m$. $\prod\left(4+x_k^2\right)=\dfrac{27625}{16}=\dfrac{125}{16}\cdot221$, so $m=221$. Answer: 221