Question

If $a+\frac{1}{a}=4$, then the value of $a^2-\frac{1}{a^2}$ is?

• A. $8\sqrt{3}$
• B. $2\sqrt{3}$
• C. $12$
• D. $4\sqrt{2}$

Answer 1

The correct option is A $8\sqrt{3}$

$(a+\frac{1}{a}{)}^2=a^2+\frac{1}{a^2}+2=16$

$(a-\frac{1}{a})=\sqrt{a^2+\frac{1}{a^2}-2}=\sqrt{16-2-2}=\sqrt{12}=2\sqrt{3}$

$(a^2-\frac{1}{a^2})=(a+\frac{1}{a})(a-\frac{1}{a})=\left(4\right)\left(2\sqrt{3}\right)=8\sqrt{3}$