JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4\) 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 \(\mathrm{m}\) is ..........
- A \(357\)
- B \(347\)
- C \(657\)
- D \(221\)
Answer & Solution
Correct Answer
(D) \(221\)
Step-by-step Solution
Detailed explanation
\( 4 x^4+8 x^3-17 x^2-12 x+9 \) \( =4\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\) Put \(x=2 i -2 i\) \( 64-64 i+68-24 i+9=\left(2 i-x_1\right)\left(2 i-x_2\right)\left(2 i-x_3\right) \) \( \left(2 \mathrm{i}-\mathrm{x}_4\right) \)…
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