Question

If $x=2+\sqrt{3}$ then find the value of $x^2+\frac{1}{x^2}$.

Answer 1

$x=2+\sqrt{3}$

To find :$x^2+\frac{1}{x^2}$

Sol : $x=2+\sqrt{3}$

$x^2=(2+\sqrt{3}{)}^2=4+3+4\sqrt{3}=7+4\sqrt{3}$

$\frac{1}{x}=\frac{1}{2+\sqrt{3}}$

$\frac{1}{x}=\frac{2-\sqrt{3}}{2-\sqrt{3}}\times \frac{1}{2+\sqrt{3}}$

$\frac{1}{x}=\frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{1}$

${\left(\frac{1}{x}\right)}^2=(2-\sqrt{3}{)}^2$

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

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

$\therefore x^2+\frac{1}{x^2}=14$