Question

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

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

Answer 1

Answer: (D)

$y^2-8y+9=0$

Given, $y=x+\frac{1}{x},x\ne 0$ and $(x^2-3x+1)(x^2-5x+1)=6x$

Thus $(x-3+\frac{1}{x})(x-5+\frac{1}{x})=6$ ( divided by $x$ )

$\Rightarrow (y-3)(y-5)=6$

$\Rightarrow y^2-8y+15=6$

$\Rightarrow y^2-8y+9=0$