Question

If $a^2-5a-1=0$, then find $a^2-\frac{1}{a^2}$

• A. $\pm 5\sqrt{29}$
• B. $\pm 2\sqrt{29}$
• C. $\pm 7\sqrt{29}$
• D. $\pm 3\sqrt{29}$

Answer 1

The correct option is A $\pm 5\sqrt{29}$

$a^2-5a-1=0$

$=>a^2-1=5a$

$=>a-\frac{1}{a}=5$

Squaring both sides, we get

$=>a^2+\frac{1}{a^2}-2a\times \frac{1}{a}=25$

$=>a^2+\frac{1}{a^2}=25+2$

$=>a^2+\frac{1}{a^2}=27$

Now,

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

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

$=27+2$

$=29$

$a+\frac{1}{a}=\pm \sqrt{29}$

Now,

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

$=(\pm \sqrt{29})\left(5\right)$

$=\pm 5\sqrt{29}$