Question

If $x$ is rational and $4\left(x^2+\frac{1}{x^2}\right)+16\left(x+\frac{1}{x}\right)-57=0,$ then the product of all possible values of $x$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

Given equation,
4 $\left(x^2+\frac{1}{x^2}\right)+16\left(x+\frac{1}{x}\right)-57=0$
Let, $x+\frac{1}{x}=y;$ $x^2+\frac{1}{x^2}=y^2-2$
$\Rightarrow$ $4y^2+16y-65=0$
$\Rightarrow$ $y=-\frac{13}{2}or\frac{5}{2}$
When, $y=\frac{5}{2}$
$x+\frac{1}{x}=\frac{5}{2}\Rightarrow x=2or\frac{1}{2}$
When, $y=-\frac{13}{2}$
$\Rightarrow$ $x+\frac{1}{x}=-13/2$
$\Rightarrow$ $2x^2+13x+2=0$
$\Rightarrow$ $x=\frac{-13\pm \sqrt{153}}{4}$
Since $x$ is rational, $x=2$ or $\frac{1}{2}$
Hence, their product is $1$ .