If $(2x^2-3x+1)(2x^2+5x+1)=9x^2$,then prove that the equation has real roots. If $(2x^2-3x+1)(2x^2+5x+1)=9x^2$,then prove that the equation has real roots.
MY attempt:
We can open and get a bi quadratic but that is two difficult to show that it has real roots.THere must be an easy way.!

First divide both members of the equation by $x^2$: 
$$(2x + \frac{1}{x} - 3)(2x + \frac{1}{x} + 5) = 9$$
With the notation $2x + \frac{1}{x} = y$ equation is obtained in $y$: $$y^2 + 2y - 24 =0$$ with roots $y_1 = -6$ and $y_2=4$. 
The equation $$2x + \frac{1}{x} = -6$$ has roots $x_1,_2 = \frac{-3\pm\sqrt{7}}{2}$;
The equation $$2x + \frac{1}{x} = 4$$ has roots $x_3,_4 = \frac{2\pm\sqrt{2}}{2}$.
