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.!
$\begingroup$
$\endgroup$
asked Jul 28, 2013 at 8:56
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
11
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.
-
-
$\begingroup$ @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 $f(c)=0$. $\endgroup$– MikasaJul 28, 2013 at 9:14
-
$\begingroup$ I see that. But how about the other roots? $\endgroup$ Jul 28, 2013 at 9:15
-
$\begingroup$ @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. $\endgroup$– MikasaJul 28, 2013 at 9:18
-
1$\begingroup$ @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. $\endgroup$ Jul 28, 2013 at 9:20
$\begingroup$
$\endgroup$
$$(2x^2-3x+1)(2x^2+5x+1)=(2x^2+x+1)^2-16x^2=9x^2\\\Rightarrow 2x^2+x+1=\pm5x\\ \Rightarrow 2x^2+ax+1=0 $$ where $a=-4\ \mbox{or}\ 6$. Hence the discriminant, $$\Delta=a^2-8>0$$ always. So the roots are real.
answered Jul 28, 2013 at 9:07
$\begingroup$
$\endgroup$
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 $x_3,_4 = \frac{2\pm\sqrt{2}}{2}$.
