Question

find the square root of the number. answer in exact form. -4\sqrt{-32}

Explanation:

Step1: Simplify the number inside the square - root

First, rewrite $\sqrt{-32}$. We know that $-32 = 32\times(-1)$, so $\sqrt{-32}=\sqrt{32}\times\sqrt{-1}$. Since $\sqrt{-1}=i$ and $\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}$, then $\sqrt{-32}=4\sqrt{2}i$. So the original number $-4\sqrt{-32}=-4\times4\sqrt{2}i=-16\sqrt{2}i$.

Step2: Let the square - root of $-16\sqrt{2}i$ be $a + bi$

$(a + bi)^2=-16\sqrt{2}i$. Expand the left - hand side: $(a + bi)^2=a^{2}-b^{2}+2abi$. So we have the system of equations

$$\begin{cases}a^{2}-b^{2}=0\\2ab=-16\sqrt{2}\end{cases}$$

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From $a^{2}-b^{2}=0$, we get $a = b$ or $a=-b$.

Case 1: If $a = b$

Substitute $a = b$ into $2ab=-16\sqrt{2}$, then $2a\times a=-16\sqrt{2}$, i.e., $2a^{2}=-16\sqrt{2}$, $a^{2}=-8\sqrt{2}$, no real solutions.

Case 2: If $a=-b$

Substitute $a=-b$ into $2ab=-16\sqrt{2}$, then $2(-b)\times b=-16\sqrt{2}$, i.e., $-2b^{2}=-16\sqrt{2}$, $b^{2}=8\sqrt{2}$.
So $b=\pm\sqrt{8\sqrt{2}}=\pm2\sqrt{2\sqrt{2}}$ and $a=\mp2\sqrt{2\sqrt{2}}$.
The square roots are $2\sqrt{2\sqrt{2}}-2\sqrt{2\sqrt{2}}i$ and $-2\sqrt{2\sqrt{2}} + 2\sqrt{2\sqrt{2}}i$.

Answer:

$2\sqrt{2\sqrt{2}}-2\sqrt{2\sqrt{2}}i,-2\sqrt{2\sqrt{2}} + 2\sqrt{2\sqrt{2}}i$