1. **State the problem:** We want to express $\sqrt{3 + 2\sqrt{2}}$ in the form $m + n\sqrt{2}$ where $m$ and $n$ are integers. 2. **Recall the formula:** Given $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2$, we want to find $a$ and $b$ such that: $$ (a + b\sqrt{2})^2 = 3 + 2\sqrt{2} $$ 3. **Set up equations by matching terms:** Equate the rational parts and the irrational parts: $$ a^2 + 2b^2 = 3 $$ $$ 2ab = 2 $$ 4. **Solve for $a$ and $b$:** From $2ab = 2$, divide both sides by 2: $$ \cancel{2}ab = \cancel{2} \Rightarrow ab = 1 $$ 5. **Express $b$ in terms of $a$:** $$ b = \frac{1}{a} $$ 6. **Substitute into the first equation:** $$ a^2 + 2\left(\frac{1}{a}\right)^2 = 3 $$ $$ a^2 + \frac{2}{a^2} = 3 $$ 7. **Multiply both sides by $a^2$ to clear denominator:** $$ a^4 + 2 = 3a^2 $$ 8. **Rearrange to form a quadratic in $a^2$:** $$ a^4 - 3a^2 + 2 = 0 $$ Let $x = a^2$, then: $$ x^2 - 3x + 2 = 0 $$ 9. **Solve quadratic equation:** $$ (x - 1)(x - 2) = 0 $$ So, $x = 1$ or $x = 2$ 10. **Find $a$ from $x = a^2$:** - If $a^2 = 1$, then $a = \pm 1$ - If $a^2 = 2$, then $a = \pm \sqrt{2}$ 11. **Find corresponding $b$ values:** - For $a = 1$, $b = 1/1 = 1$ - For $a = -1$, $b = 1/(-1) = -1$ - For $a = \sqrt{2}$, $b = 1/\sqrt{2} = \frac{\sqrt{2}}{2}$ (not integer) - For $a = -\sqrt{2}$, $b = -\frac{\sqrt{2}}{2}$ (not integer) 12. **Choose integer solutions:** $a = 1$, $b = 1$ 13. **Write the final answer:** $$ \sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2} $$ This matches the original expression when squared.