Question

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

Answer 1

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

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

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