The number of real roots of the equation $(x^{2} + 2)^{2} + 8 x^{2} = 6 x (x^{2} + 2)$ is/are

0

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2

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4

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6

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Solution

The correct option is A $2$
Given equation is $(x^{2} + 2)^{2} + 8 x^{2} = 6 x (x^{2} + 2)$
${(\frac{x^{2} + 2}{x})}^{2} + 8 = 6 (\frac{x^{2} + 2}{x})$
Substitute $\frac{x^{2} + 2}{x} = y$
$\Rightarrow y^{2} - 6 y + 8 = 0$
$\Rightarrow (y - 2) (y - 4) = 0$ $\Rightarrow y = 2, 4$
When $y = 2 \Rightarrow \frac{x^{2} + 2}{x} = 2$
$\Rightarrow x^{2} - 2 x + 2 = 0$ $\Rightarrow x = 1 \pm i$
When $y = 4 \Rightarrow \frac{x^{2} + 2}{x} = 4$
$\Rightarrow x^{2} - 4 x + 2 = 0$
$\Rightarrow x = 2 \pm \sqrt{2}$
Therefore, two real roots. Hence, option 'B' is correct.

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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

R

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

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