Question

If $a-\frac{1}{a}=8$ and $a\ne 0$, find $a^2-\frac{1}{a^2}$.

• A. $\pm 11\sqrt{17}$
• B. $\pm 13\sqrt{17}$
• C. $\pm 16\sqrt{17}$
• D. $\pm 17\sqrt{17}$

Answer 1

The correct option is C $\pm 16\sqrt{17}$

$a-\frac{1}{a}=8$

Squaring both sides,

$=(a-\frac{1}{a}{)}^2=8^2$

$=a^2+(\frac{1}{a}{)}^2-2(a\left)\right(\frac{1}{a})=64$

$=a^2+\frac{1}{a^2}-2=64$

$=a^2+\frac{1}{a^2}=66$

Now,

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

$=66+2$

$=68$

$=>a+\frac{1}{a}=\sqrt{68}$

$=\pm 2\sqrt{1}7$

Now,

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

$=\pm \left(2\sqrt{1}7\right)\left(8\right)$

$=\pm 16\sqrt{1}7$