Let x 1 , x 2 , x 3 , x 4 be the solutions of the equation 4 x 4 + 8 x 3 - 17 x 2 - 12 x + 9 = 0 and ( 4 + x 1 2 ) ( 4 + x 2 2 ) ( 4 + x 3 2 ) ( 4 + x 4 2 ) = 125 16 m . Then the value of m is _______ . [2024]

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Solution

Q.
Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solutions of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m .$ Then the value of $m$ is _______ . [2024]

Ans.
(221)

We have, $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are solution of $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$

$\Rightarrow 4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9$

$= 4 (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4})$ ...(i)

Let us substitute $x = 2 i$ in (i)

$∴ 64 - 64 i + 68 - 24 i + 9$

$= 4 (2 i - x_{1}) (2 i - x_{2}) (2 i - x_{3}) (2 i - x_{4})$

$\Rightarrow 4 (2 i - x_{1}) (2 i - x_{2}) (2 i - x_{3}) (2 i - x_{4}) = 141 - 88 i$ ...(ii)

Substitute $x = - 2 i$ in (i), we get

$4 (2 i + x_{1}) (2 i + x_{2}) (2 i + x_{3}) (2 i + x_{4}) = 141 + 88 i$ ...(iii)

Multiplying (ii) and (iii), we get

$(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{{(141)}^{2} + 88^{2}}{16}$

$\Rightarrow \frac{125 m}{16} = \frac{27625}{16} \Rightarrow m = 221$

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