Question

If $x^2+\frac{1}{x^2}=102$, then find the value of $\frac{x^2-5x-1}{x}$

• A. 15
• B. -5
• C. 10
• D. -10

Answer 1

Answer: (A), (B)

$x^2+\frac{1}{x^2}=102$
Subtract 2 on both sides:
$x^2+\frac{1}{x^2}-2=102-2$
$\Rightarrow {\left(x-\frac{1}{x}\right)}^2=100\Rightarrow x-\frac{1}{x}=\pm 10$
$\Rightarrow \frac{x^2+5x-1}{x}=\frac{\frac{x^2}{x}-\frac{5}{x}+\frac{1}{x}}{\frac{x}{x}}$
$\{$ Divide the numerator and denominator by $x\}$
$\Rightarrow \frac{\left(x-\frac{1}{x}\right)+5}{1}$
When $x-\frac{1}{x}=10$
Then $\frac{10+5}{1}=15$
When $x-\frac{1}{x}=-10\Rightarrow \frac{-10+5}{1}=-5$
Possible values of $x=15,-5$