Question

Solve: If $x=5+2\sqrt{6}$, find $\frac{1}{x},(x^2+\frac{1}{x^2})$

Answer 1

Given:

$x=5–2\sqrt{6}$

so,

$1/x=1/(5–2\sqrt{6})$

By rationalizing the denominator,

$1/(5–2\sqrt{6})=\left[1\right(5+2\sqrt{6}\left)\right]/\left[\right(5–2\sqrt{6}\left)\right(5+2\sqrt{6}\left)\right]$

$=\left[\right(5+2\sqrt{6}\left)\right]/[5^2–(2\sqrt{6}{)}^2]$

$=\left[\right(5+2\sqrt{6}\left)\right]/[25–4.6]$

$=\left[\right(5+2\sqrt{6}\left)\right]/[25–24]$

$=(5+2\sqrt{6})$

Then,

$x+1/x=5–2\sqrt{6}+(5+2\sqrt{6})$

$=10$

Square on both sides we get

$(x+1/x{)}^2={10}^2$

$x^2+1/x^2+2x.1/x=100$

$x^2+1/x^2+2=100$

$x^2+1/x^2=100–2$

$=98$