Question

The product of all the real roots of $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ is

• A. $2$
• B. $1$
• C. $3$
• D. $7$

Answer 1

The correct option is B $1$

Given equation $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$

or $(x^2+\frac{1}{x^2})-8(x+\frac{1}{x})+9=0$

or ${(x+\frac{1}{x})}^2-2-8(x+\frac{1}{x})+9=0$

or ${(x+\frac{1}{x})}^2-8(x+\frac{1}{x})+7=0$

Substituting $x+\frac{1}{x}=t$ we get

$t^2-8t+7=0⟹(t-7)(t-1)=0$

$⟹t=7\text{or}t=1$

$⟹x+\frac{1}{x}=7\text{or}x+\frac{1}{x}=1$

$⟹x^2-7x+1=0\text{or}{x}^2-x+1=0$

$⟹x=\frac{7\pm \sqrt{49-4}}{2}\text{or}x=\frac{1\pm \sqrt{1-4}}{2}$

$⟹x=\frac{7\pm \sqrt{45}}{2}orx=\frac{1\pm i\sqrt{3}}{2}$

$\therefore$ Roots of the given equation are $\frac{7\pm \sqrt{45}}{2},\frac{1\pm i\sqrt{3}}{2}$

$\therefore$Product of the roots

$=\frac{7+\sqrt{45}}{2}\times \frac{7-\sqrt{45}}{2}\times \frac{1+i\sqrt{3}}{2}\times \frac{1-i\sqrt{3}}{2}=\frac{49-45}{4}\times \frac{1+3}{4}=1\times 1=1$