The number of distinct positive real roots of the equation $(x^{2} + 6)^{2} - 35 x^{2} = 2 x (x^{2} + 6)$ is

4

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3

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2

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0

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Solution

The correct option is C $2$
$(x^{2} + 6)^{2} - 35 x^{2} = 2 x (x^{2} + 6)$
$\Rightarrow \frac{x^{2} + 6}{x} - 35 (\frac{x}{x^{2} + 6}) = 2$
Let $\frac{x^{2} + 6}{x} = t$
Then, $t - \frac{35}{t} = 2$
$\Rightarrow t^{2} - 2 t - 35 = 0$
$\Rightarrow (t - 7) (t + 5) = 0$
$\Rightarrow t = 7$ or $t = - 5$
$\Rightarrow x^{2} - 7 x + 6 = 0$ or $x^{2} + 5 x + 6 = 0$
$\Rightarrow x = 1, 6$ or $x = - 2, - 3$
$∴$ Roots of given equations are $- 3, - 2, 1, 6$
Hence, the number of distinct positive real roots is $2$ .

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