4-2.Quadratic Equations and Inequations
hard

ધારોકે $x_1, x_2, x_3, x_4$ એ સમીકરણ $4 x^4+8 x^3-17 x^2-12 x+9=0$ નાં બીજ છે અને $\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$. તો $m$ નું મૂલ્ય ............ છે. 

A

$357$

B

$347$

C

$657$

D

$221$

(JEE MAIN-2024)

Solution

$ 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) $

$ =141-88 \mathrm{i} $                  …………….($1$)

$ 64+64 \mathrm{i}+68+24 \mathrm{i}+9=4\left(-2 \mathrm{i}-\mathrm{x}_1\right)\left(-2 \mathrm{i}-\mathrm{x}_2\right)(-2 \mathrm{i} $

$ \left.-\mathrm{x}_3\right)\left(-2 \mathrm{i}-\mathrm{x}_4\right) $

$ =141+88 \mathrm{i} $                  …………………($2$)

$ \frac{125}{16} \mathrm{~m}=\frac{141^2+88^2}{16} $

$ \mathrm{~m}=221$

Standard 11
Mathematics

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