Question

If $a-\frac{1}{a}=8$ and $a\ne 0$, then find $a+\frac{1}{a}$

• A. $\pm 2\sqrt{17}$
• B. $\pm 2\sqrt{15}$
• C. $\pm 2\sqrt{19}$
• D. $\pm 2\sqrt{11}$

Answer 1

The correct option is A $\pm 2\sqrt{17}$

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

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,

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

$=66+2$

$=68$

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

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