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

Step-by-step solution

1. Given $x=3+2\sqrt{2}$, find $\frac{1}{x}=\frac{1}{3+2\sqrt{2}}$.
2. Rationalise $\frac{1}{x}$ by multiplying numerator and denominator by $(3-2\sqrt{2})$: $\frac{1}{x}=\frac{3-2\sqrt{2}}{(3+2\sqrt{2})(3-2\sqrt{2})}=\frac{3-2\sqrt{2}}{3^2-(2\sqrt{2}{)}^2}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}$.
3. Add $x$ and $\frac{1}{x}$: $x+\frac{1}{x}=(3+2\sqrt{2})+(3-2\sqrt{2})=6$.
4. Square both sides of $x+\frac{1}{x}=6$: ${\left(x+\frac{1}{x}\right)}^2=6^2⟹x^2+\frac{1}{x^2}+2\cdot x\cdot \frac{1}{x}=36$.
5. Simplify the equation to solve for $x^2+\frac{1}{x^2}$: $x^2+\frac{1}{x^2}+2=36⟹x^2+\frac{1}{x^2}=36-2=34$.

Concept tested: Operations on real numbers and algebraic identities

Published 20 August 2026 · How this solution was prepared and checked

Related questions

• 1 Simplify the expression by rationalising the denominator: 5/(√7 - √2).
• 2 Express 0.6 in the form p/q, where p and q are integers and q ≠ 0.