1. **State the problem:** Find the square root of the expression $17 + 12\sqrt{2}$. 2. **Recall the formula:** To find the square root of a number in the form $a + b\sqrt{c}$, we look for numbers $x$ and $y$ such that: $$\sqrt{a + b\sqrt{c}} = x + y\sqrt{c}$$ Squaring both sides gives: $$a + b\sqrt{c} = (x + y\sqrt{c})^2 = x^2 + 2xy\sqrt{c} + y^2 c$$ This means: $$a = x^2 + y^2 c$$ $$b = 2xy$$ 3. **Apply to our problem:** Here, $a = 17$, $b = 12$, and $c = 2$. We need to find $x$ and $y$ such that: $$17 = x^2 + 2 y^2$$ $$12 = 2 x y$$ From the second equation: $$xy = 6$$ 4. **Solve the system:** From $xy = 6$, express $y = \frac{6}{x}$. Substitute into the first equation: $$17 = x^2 + 2 \left(\frac{6}{x}\right)^2 = x^2 + 2 \frac{36}{x^2} = x^2 + \frac{72}{x^2}$$ Multiply both sides by $x^2$: $$17 x^2 = x^4 + 72$$ Rearranged: $$x^4 - 17 x^2 + 72 = 0$$ Let $z = x^2$, then: $$z^2 - 17 z + 72 = 0$$ 5. **Solve quadratic in $z$:** $$z = \frac{17 \pm \sqrt{17^2 - 4 \times 72}}{2} = \frac{17 \pm \sqrt{289 - 288}}{2} = \frac{17 \pm 1}{2}$$ So: $$z_1 = \frac{18}{2} = 9, \quad z_2 = \frac{16}{2} = 8$$ 6. **Find $x$ and $y$:** If $x^2 = 9$, then $x = 3$ or $-3$. Corresponding $y = \frac{6}{x} = 2$ or $-2$. Check if these satisfy the first equation: $$x^2 + 2 y^2 = 9 + 2 \times 4 = 9 + 8 = 17$$ Correct. 7. **Final answer:** $$\sqrt{17 + 12\sqrt{2}} = 3 + 2\sqrt{2}$$ This is the positive root since square roots are non-negative.