Question

If $a^2+\frac{1}{a^2}=47$ and $a\ne 0$, find $a+\frac{1}{a}$.

• A. $\pm 6$
• B. $\pm 12$
• C. $\pm 7$
• D. $\pm 9$

Answer 1

The correct option is C $\pm 7$

Given $a^2+\frac{1}{a^2}=47$

Now, $(a+\frac{1}{a}{)}^2=a^2+\frac{1}{a^2}+2a(\frac{1}{a})$

$=>(a+\frac{1}{a}{)}^2=47+2$

$=>a+\frac{1}{a}=\sqrt{49}$

$=>a+\frac{1}{a}=\pm 7$