Question

If $x=2+\sqrt{3}$, Find $x^2+\frac{1}{x^2}?$

Answer 1

It is given that $x=2+\sqrt{3}$. We find $x^2$ and $\frac{1}{x^2}$ as follows:

$x^2=(2+\sqrt{3}{)}^2=(2{)}^2+(\sqrt{3}{)}^2+(2\times 2\times \sqrt{3}\left)\right(\because (a+b{)}^2=a^2+b^2+2ab)=4+3+4\sqrt{3}=7+4\sqrt{3}$

$\frac{1}{x^2}=\frac{1}{7+4\sqrt{3}}=\frac{1}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}\left(Rationalizing\right)=\frac{7-4\sqrt{3}}{(7+4\sqrt{3})(7-4\sqrt{3})}=\frac{7-4\sqrt{3}}{7^2-(4\sqrt{3}{)}^2}(\because (a+b\left)\right(a-b)=a^2-b^2)$

$=\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$

Now, we calculate $x^2+\frac{1}{x^2}$ as shown below:

$x^2+\frac{1}{x^2}=7+4\sqrt{3}+(7-4\sqrt{3})=7+4\sqrt{3}+7-4\sqrt{3}=14$

Hence, $x^2+\frac{1}{x^2}=14$.