If $(2x^2-3x+1)(2x^2+5x+1)=9x^2$,then prove that the equation has real roots.

Question

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.!

Answer 1

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 $x_3,_4 = \frac{2\pm\sqrt{2}}{2}$.

Answer 2

Hint: Note that if we set $$f(x)=(2x^2-3x+1)(2x^2+5x+1)-9x^2$$ then $$f(1)\times f(2)<0$$ Now use Intermediate value theorem.

Answer 3

$$(2x^2-3x+1)(2x^2+5x+1)=(2x^2+x+1)^2-16x^2=9x^2\\\Rightarrow 2x^2+x+1=\pm5x\\ \Rightarrow 2x^2+ax+1=0 $$ where $a=-4\ \mbox{or}\ 6$. Hence the discriminant, $$\Delta=a^2-8>0$$ always. So the roots are real.