1. **Problem statement:** Find the two possible values of $\sqrt{-5 - 12i}$ in the form $a + bi$, where $a,b \in \mathbb{R}$. 2. **Formula and approach:** We want to find complex numbers $z = a + bi$ such that: $$z^2 = -5 - 12i$$ Expanding $z^2$: $$ (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 - b^2 + 2abi $$ Equate real and imaginary parts: $$ a^2 - b^2 = -5 \quad \text{(real part)} $$ $$ 2ab = -12 \quad \text{(imaginary part)} $$ 3. **Solve the system:** From the imaginary part: $$ 2ab = -12 \implies ab = -6 \implies b = \frac{-6}{a} $$ Substitute into the real part: $$ a^2 - \left(\frac{-6}{a}\right)^2 = -5 $$ $$ a^2 - \frac{36}{a^2} = -5 $$ Multiply both sides by $a^2$: $$ a^4 - 36 = -5a^2 $$ Rearranged: $$ a^4 + 5a^2 - 36 = 0 $$ 4. **Let $x = a^2$, solve quadratic:** $$ x^2 + 5x - 36 = 0 $$ Use quadratic formula: $$ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-36)}}{2} = \frac{-5 \pm \sqrt{25 + 144}}{2} = \frac{-5 \pm \sqrt{169}}{2} $$ $$ x = \frac{-5 \pm 13}{2} $$ Two solutions: $$ x_1 = \frac{-5 + 13}{2} = 4 $$ $$ x_2 = \frac{-5 - 13}{2} = -9 $$ Since $a^2 = x$, $a^2$ must be real and non-negative, so $a^2 = 4$. 5. **Find $a$ and $b$:** $$ a = \pm 2 $$ Recall $b = \frac{-6}{a}$: - If $a = 2$, then $b = \frac{-6}{2} = -3$ - If $a = -2$, then $b = \frac{-6}{-2} = 3$ 6. **Check solutions:** $$ (2 - 3i)^2 = 4 - 12i + 9i^2 = 4 - 12i - 9 = -5 - 12i $$ $$ (-2 + 3i)^2 = 4 - 12i + 9i^2 = 4 - 12i - 9 = -5 - 12i $$ Both satisfy the equation. **Final answer:** $$ \boxed{2 - 3i \quad \text{and} \quad -2 + 3i} $$