Question

If $y=x+\frac{1}{x},x\ne 0$, then the equation $\left(x^2-3x+1\right)\left(x^2-5x+1\right)=6x^2$ reduces to

• A. $y^2-8y+7=0$
• B. $y^2+8y+7=0$
• C. $y^2-8y-9=0$
• D. $y^2-8y+9=0$
• E. $y^2-7y+13=0$

Answer 1

Answer: (D)

Given, $y=x+\frac{1}{x}$
and $\left(x^2-3x+1\right)\left(x^2-5x+1\right)=6x$
$\Rightarrow \left(x-3+\frac{1}{x}\right)\left(x-5+\frac{1}{x}\right)=6($ divide by $x)$
$\Rightarrow (y-3)(y-5)=6$
$\Rightarrow y^2-8y+15=6$
$\Rightarrow y^2-8y+9=0$