Question

If $x=\frac{3+\sqrt{7}}{2}$, find the value of $4x^2+\frac{1}{x^2}$

Answer 1

$x=\frac{3+\sqrt{7}}{2}$

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

$=(3+\sqrt{7}{)}^2+\frac{4\ast (3-\sqrt{7}{)}^2}{(9-7{)}^2}$

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

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

$=\left(9\right)+(\sqrt{7}{)}^2$

$=16$