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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.!

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up vote 1 down vote accepted

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

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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.

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+1 Although I didn't get it – Mahdi Khosravi Jul 28 '13 at 9:10
@MahdiKhosravi: $f$ is continuous on $[1,2]$ and $f(1)$ and $f(2)$ have the opposite signs so there is a $c\in(1,2)$ such that $f(c)=0$. – Babak S. Jul 28 '13 at 9:14
I see that. But how about the other roots? – Mahdi Khosravi Jul 28 '13 at 9:15
@MahdiKhosravi: Maybe the OP can change the intervals. I say this because the structure of the function is not that complicated than the OP couldn't examine other numbers. – Babak S. Jul 28 '13 at 9:18
@SamratMukhopadhyay But the degree is four, finding one real guarantees one other real root. Two roots remain and they can be both real or both complex conjugate. – Mahdi Khosravi Jul 28 '13 at 9:20

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

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