Question

If $a+\frac{1}{a}=6$ and $a\ne 0$; find $a^2-\frac{1}{a^2}$ .

• A. $\pm 24\sqrt{3}$
• B. $\pm 4\sqrt{2}$
• C. $\pm 24\sqrt{2}$
• D. $\pm 4\sqrt{3}$

Answer 1

Answer: (C)

$\pm 24\sqrt{2}$

$a+\frac{1}{a}=6$
Squaring both sides,
$=(a+\frac{1}{a}{)}^2=6^2$
$=a^2+(\frac{1}{a}{)}^2+2(a\left)\right(\frac{1}{a})=36$
$=a^2+\frac{1}{a^2}+2=36$
$=a^2+\frac{1}{a^2}=34$
Now,
$(a-\frac{1}{a}{)}^2=a^2+\frac{1}{a^2}-2(a\left)\right(\frac{1}{a})$
$=34-2$
$=32$
$=>a-\frac{1}{a}=\pm \sqrt{32}$
$=\pm 4\sqrt{2}$
Now,
$(a^2-\frac{1}{a^2})=(a+\frac{1}{a})(a-\frac{1}{a})$
$=6(\pm 4\sqrt{2})$
$=\pm 24\sqrt{2}$