Question

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

Answer 1

If x=$3-2\sqrt{2}$ , then $\frac{1}{x}=\frac{1}{3-2\sqrt{2}}$

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

$\Rightarrow \frac{3+2\sqrt{2}}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}$ = $\frac{3+2\sqrt{2}}{9-8}$ = $3+2\sqrt{2}$

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

Now, $x^2+\frac{1}{x^2}$ = ${(3-2\sqrt{2})}^2+{(3+2\sqrt{2})}^2$

Using ${(a+b)}^2=a^2+2ab+b^2,{(a-b)}^2=a^2-2ab+b^2$

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

= $(9-12\sqrt{2}+8)+(9+12\sqrt{2}+8)$

= $17-12\sqrt{2}+17+12\sqrt{2}$

= $34$

Hence, the value of $x^2+\frac{1}{x^2}$ = $34$

1x=3+22(9−8)=3+22\dfrac{1}{x}=\dfrac{3+2\sqrt { 2 }} {(9-8)}=3+2\sqrt { 2 }