Question

If $\left(2x^2-3x+1\right)\left(2x^2+5x+1\right)=9x^2$, then equation has

• A. four real roots
• B. two real and two imaginary roots
• C. four imaginary roots
• D. None of the above

Answer 1

The correct option is A four real roots

$(2x^2-3x+1)(2x^2+5x+1)=9x^2$

$\frac{(2x^2-3x+1)(2x^2+5x+1)}{x^2}=9$

$\frac{(2x^2-3x+1)}{x}\frac{(2x^2+5x+1)}{x}=9$

$(2x-3+\frac{1}{x})(2x+5+\frac{1}{x})=9$

$(2x+\frac{1}{x}-3)(2x+\frac{1}{x}+5)=9$

Let $y=2x+\frac{1}{x}$

$\Rightarrow (y-3)(y+5)=9$

$\Rightarrow y^2+2y-15=9$

$\Rightarrow y^2+2y-15-9=0$

$\Rightarrow y^2+2y-24=0$

$\Rightarrow (y+6)(y-4)=0$

$\Rightarrow (2x+\frac{1}{x}+6)(2x+\frac{1}{x}-4)=0$ where $y=2x+\frac{1}{x}$

$\Rightarrow (2x^2+1+6x)(2x^2+1-4x)=0$

$\Rightarrow 2x^2+6x+1=0,2x^2-4x+1=0$

$\therefore x=\frac{-6\pm \sqrt{36-4\times 2\times 1}}{2\times 2},x=\frac{4\pm \sqrt{16-4\times 2\times 1}}{2\times 2}$

$\therefore x=\frac{-6\pm \sqrt{36-8}}{4},x=\frac{4\pm \sqrt{16-8}}{4}$

$\therefore x=\frac{-6\pm \sqrt{28}}{4},x=\frac{4\pm \sqrt{8}}{4}$

$\therefore x=\frac{-6\pm 2\sqrt{7}}{4},x=\frac{4\pm 2\sqrt{2}}{4}$

$\therefore x=\frac{-3\pm \sqrt{7}}{2},x=\frac{2\pm \sqrt{2}}{2}$

$\therefore$ Roots are $\{\frac{-3+\sqrt{7}}{2},\frac{-3-\sqrt{7}}{2},\frac{2+\sqrt{2}}{2},\frac{2-\sqrt{2}}{2}\}$

Thus, there are four real roots.