If $(2 x^{2} - 3 x + 1) (2 x^{2} + 5 x + 1) = 9 x^{2}$ , then the absolute value of the sum of all real roots of the equation is

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Solution

$(2 x^{2} - 3 x + 1) (2 x^{2} + 5 x + 1) = 9 x^{2}$
As $x = 0$ is not a root of the equation, dividing the equation by $x^{2}$ ,
$(2 x + \frac{1}{x} - 3) (2 x + \frac{1}{x} + 5) = 9$
Assuming $2 x + \frac{1}{x} = y$ ,
$(y - 3) (y + 5) = 9 \Rightarrow y^{2} + 2 y - 24 = 0 \Rightarrow (y + 6) (y - 4) = 0 \Rightarrow y = - 6, 4$

Case $1 :$ When $y = - 6$
$2 x + \frac{1}{x} = - 6 \Rightarrow 2 x^{2} + 6 x + 1 = 0 D = 36 - 8 = 28 > 0$
Sum of real roots $= - \frac{6}{2} = - 3$
Case $2 :$ When $y = 4$
$2 x + \frac{1}{x} = 4 \Rightarrow 2 x^{2} - 4 x + 1 = 0 D = 16 - 8 = 8 > 0$
Sum of real roots $= 2$

Hence, the absolute value of the sum of all real roots is $1.$

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