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

$2$

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None

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$4$

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$0$

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Solution

The correct option is A
$2$

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

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

$\Rightarrow (x^{4} + 4 x^{2} + 4) - 6 x^{3} - 12 x + 8 x^{2} = 0$

$\Rightarrow x^{4} - 6 x^{3} + 12 x^{2} - 12 x + 4 = 0$

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

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

Calculating the discriminant value for both the equations.

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

$D = b^{2} - 4 a c = 16 - 4 (1) (2) = 8 > 0$

$x^{2} - 2 x + 2 = 0$

$D = b^{2} - 4 a c = 4 - 4 (1) (2) = - 4 < 0$

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

Hence, the number of real roots are $2$ .

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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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