The sum of real roots of the equation $(x - 1) (x - 2) (3 x - 1) (3 x - 2) = 8 x^{2}$ is

3

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4

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6

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8

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Solution

The correct option is A $3$
$(x - 1) (x - 2) (3 x - 1) (3 x - 2) = 8 x^{2} \Rightarrow (x - 1) (3 x - 2) (x - 2) (3 x - 1) = 8 x^{2} \Rightarrow (3 x^{2} - 5 x + 2) (3 x^{2} - 7 x + 2) = 8 x^{2} \Rightarrow (3 x + \frac{2}{x} - 5) (3 x + \frac{2}{x} - 7) = 8$
Assuming $3 x + \frac{2}{x} - 5 = t$
$\Rightarrow t (t - 2) = 8 \Rightarrow t^{2} - 2 t - 8 = 0 \Rightarrow (t - 4) (t + 2) = 0 \Rightarrow t = - 2, 4$

When $t = - 2$
$3 x + \frac{2}{x} - 5 = - 2 \Rightarrow 3 x^{2} - 3 x + 2 = 0 D = 9 - 24 < 0$
No real roots.

When $t = 4$
$3 x + \frac{2}{x} - 5 = 4 \Rightarrow 3 x^{2} - 9 x + 2 = 0 D = 81 - 24 > 0$
Real and distinct roots.

Therefore,
The sum of real roots $= \frac{9}{3} = 3$

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

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)

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