The number of real solutions of $(x - 1) (x + 1) (2 x + 1) (2 x - 3) = 15$ is

2

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4

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3

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0

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Solution

The correct option is A $2$
$(x - 1) (x + 1) (2 x + 1) (2 x - 3) = 15 \Rightarrow (x - 1) (2 x + 1) (x + 1) (2 x - 3) = 15$
$\Rightarrow (2 x^{2} - x - 1) (2 x^{2} - x - 3) = 15$
Assuming $2 x^{2} - x = y$ ,
$(y - 1) (y - 3) = 15 \Rightarrow y^{2} - 4 y + 3 = 15$
$\Rightarrow y^{2} - 4 y - 12 = 0 \Rightarrow (y - 6) (y + 2) = 0$
$\Rightarrow y = 6, - 2$
Now,
when $y = 6$
$2 x^{2} - x = 6 \Rightarrow 2 x^{2} - x - 6 = 0 \Rightarrow (2 x + 3) (x - 2) = 0$
$\Rightarrow x = - \frac{3}{2}, 2$
and
when $y = - 2$
$2 x^{2} - x = - 2 \Rightarrow 2 x^{2} - x + 2 = 0$
$D = 1 - 4 (2) (2) < 0$
No real roots.

Hence, the equation has $2$ real roots.

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