${(x^{2} + 2)}^{2} + 8 x^{2} = 6 x (x^{2} + 2)$ , then the sum of real roots is

2 \sqrt{2}

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4

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Solution

The correct option is C $4$
$(x^{2} + 2)^{2} + 8 x^{2} = 6 x (x^{2} + 2)$

Let $x^{2} + 2 = t$

$t^{2} + 8 (t - 2) = 6 x (t)$

$t^{2} - 6 x t + 8 x^{2} = 0$

$t^{2} - 4 x t - 2 x t + 8 x^{2} = 0$

$t (t - 4 x) - 2 x (t - 4 x) = 0$

$(t - 2 x) (t - 4 x) = 0$

$(x^{2} - 2 x + 2) (x^{2} - 4 x + 2) = 0$

$(\frac{2 \pm \sqrt{4 - 4 (2) (1)}}{2 (1)})$ $(\frac{4 \pm \sqrt{16 - 4 (2) (1)}}{2 (1)})$

$∴$ it has only imagining roots $\frac{4 \pm \sqrt{8}}{2} = 2 \pm 2 \sqrt{2}$

Sum of real roots $= (2 + 2 \sqrt{2}) + (2 - 2 \sqrt{2})$

$= 4$

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Similar questions

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)

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

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

i \pm k

k .

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