Question

If $a^2+\frac{1}{a^2}=18$ where $a\ne 0$, find $a-\frac{1}{a}$.

• A. $\pm 7$
• B. $-9$
• C. $\pm 4$
• D. $-1$

Answer 1

The correct option is C $\pm 4$

$a^2+\frac{1}{a^2}=18$

We know,

$(a-\frac{1}{a}{)}^2=a^2-2(a\left)\right(\frac{1}{a})+(\frac{1}{a}{)}^2$

$=>(a-\frac{1}{a}{)}^2=18-2$

$=>(a-\frac{1}{a}{)}^2=16$

$=>(a-\frac{1}{a})=\pm \sqrt{16}$

$=>(a-\frac{1}{a})=\pm 4$