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

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Solution

The correct option is B
2

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

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

$\Rightarrow x^{2} - 4 x + 2$ = 0 or $x^{2} - 2 x + 2 = 0$

Only $x^{2} - 4 x + 2 = 0$ has real roots

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Similar questions

{(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)

R

{(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

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