The sum of real solutions of the equation $(x^{2} + 2)^{2} + 8 x^{2} = 6 x (x^{2} + 2)$ , given that $x \neq 0,$ is

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Solution

The correct option is A
4

Given : $(x^{2} + 2)^{2} + 8 x^{2} = 6 x (x^{2} + 2)$

${(x + \frac{2}{x})}^{2} + 8 = 6 (x + \frac{2}{x})$

Let $x + \frac{2}{x} = t$
$\Rightarrow t^{2} - 6 t + 8 = 0$
$\Rightarrow t = 4$ or $t = 2$
$\Rightarrow x + \frac{2}{x} = 4$

or $x + \frac{2}{x} = 2$ (no real solutions)
$\Rightarrow x^{2} - 4 x + 2 = 0$
$\Rightarrow x = 2 \pm \sqrt{2}$
Hence, sum of the solutions $= 2 + \sqrt{2} + 2 - \sqrt{2} = 4$

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Similar questions

{(x^{2} + 2)}^{2} + 8 x^{2} = 6 x (x^{2} + 2)

Number of real roots of the equation $(x^{2} + 2)^{2} + 8 x^{2}$ = $6 x (x^{2} + 2)$ is :

{(x^{2} + 2)}^{2} + 8 x^{2} = 6 x (x^{2} + 2)

(x^{2} + 2)^{2} + 8 x^{2} = 6 x (x^{2} + 2)

{(x^{2} + 2)}^{2} + 8 x^{2} = 6 x (x^{2} + 2)

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