Question

find the square root of the number.
-3√-49
the square root is \boxed{}

Explanation:

Step1: Simplify the square root of -49

Recall that \(\sqrt{-a}=\sqrt{a}\cdot i\) where \(i\) is the imaginary unit and \(a>0\). So, \(\sqrt{-49}=\sqrt{49}\cdot i = 7i\).

Step2: Multiply by -3

Now, we have \(-3\sqrt{-49}=-3\times7i=-21i\). But wait, the problem says "Find the square root of the number" where the number is \(-3\sqrt{-49}=-21i\)? Wait, maybe there is a misinterpretation. Wait, maybe the original expression is to simplify \(-3\sqrt{-49}\) first. Wait, let's re - examine.

Wait, if we first simplify \(\sqrt{-49}\):
We know that for a negative number \(n=-m\) (\(m > 0\)), \(\sqrt{n}=\sqrt{m}\times i\), where \(i\) is the imaginary unit with \(i^{2}=-1\). So \(\sqrt{-49}=\sqrt{49}\times i = 7i\).

Then, \(-3\sqrt{-49}=-3\times7i=-21i\). But if we want to find the square root of \(-21i\), we can use the formula for the square root of a complex number \(z = r(\cos\theta+i\sin\theta)\), the square roots are \(\pm\sqrt{r}(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})\).

First, write \(-21i\) in polar form. The modulus \(r=\vert - 21i\vert=21\), and the argument \(\theta = \frac{3\pi}{2}\) (since it lies on the negative imaginary axis).

Then, the square roots are \(\sqrt{21}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\) and \(\sqrt{21}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\)

\(\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}\), \(\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}\), so the first square root is \(\sqrt{21}(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=-\sqrt{\frac{21\times2}{4}}+i\sqrt{\frac{21\times2}{4}}=-\frac{\sqrt{42}}{2}+i\frac{\sqrt{42}}{2}\)

\(\cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2}\), \(\sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}\), so the second square root is \(\sqrt{21}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=\frac{\sqrt{42}}{2}-i\frac{\sqrt{42}}{2}\)

But maybe there is a mistake in the problem statement. Wait, maybe the original expression is to simplify \(-3\sqrt{-49}\) first. Let's re - check the problem. The problem says "Find the square root of the number. \(-3\sqrt{-49}\)".

First, simplify \(-3\sqrt{-49}\):

\(\sqrt{-49}=\sqrt{49}\times i = 7i\) (because \(i^{2}=-1\), so \((7i)^{2}=49i^{2}=-49\))

Then \(-3\sqrt{-49}=-3\times7i=-21i\)

Now, to find the square root of \(-21i\), let \(z = x + yi\) be a square root of \(-21i\), so \(z^{2}=(x + yi)^{2}=x^{2}-y^{2}+2xyi=-21i\)

We get the system of equations:
\(

$$\begin{cases}x^{2}-y^{2}=0\\2xy=-21\end{cases}$$

\)

From \(x^{2}-y^{2}=0\), we have \(x^{2}=y^{2}\), so \(x = y\) or \(x=-y\)

Case 1: If \(x = y\), then \(2x\times x=-21\Rightarrow2x^{2}=-21\), which has no real solutions.

Case 2: If \(x=-y\), then \(2x\times(-x)=-21\Rightarrow - 2x^{2}=-21\Rightarrow x^{2}=\frac{21}{2}\Rightarrow x=\pm\sqrt{\frac{21}{2}}=\pm\frac{\sqrt{42}}{2}\)

If \(x=\frac{\sqrt{42}}{2}\), then \(y =-\frac{\sqrt{42}}{2}\)

If \(x =-\frac{\sqrt{42}}{2}\), then \(y=\frac{\sqrt{42}}{2}\)

So the square roots of \(-21i\) are \(\frac{\sqrt{42}}{2}-\frac{\sqrt{42}}{2}i\) and \(-\frac{\sqrt{42}}{2}+\frac{\sqrt{42}}{2}i\)

But maybe there was a mis - reading of the problem. If the problem was just to simplify \(-3\sqrt{-49}\) (not find the square root of it), then \(-3\sqrt{-49}=-21i\). But the problem says "Find the square root of the number" where the number is \(-3\sqrt{-49}\).

Wait, maybe the original problem was written incorrectly, and it's supposed to be simplify \(-3\sqrt{-49}\). Let's assume that. Then:

\(\sqrt{-49}=7i\) (since \(i^{2}=-1\), so \((7i)^{2}=49i^{2}=-49\))

Then \(-3\times7i=-21i\)

Answer:

If we assume the problem is to simplify \(-3\sqrt{-49}\) (maybe a misstatement of "simplify" instead of "find the square root of the number"), the result is \(-21i\). If we really need to find the square root of \(-21i\), the square roots are \(\pm(\frac{\sqrt{42}}{2}-\frac{\sqrt{42}}{2}i)\) (or \(\pm\frac{\sqrt{42}}{2}(1 - i)\))