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