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 for a negative number \( a \), \( \sqrt{-a} = i\sqrt{a} \) where \( i \) is the imaginary unit with \( i^2=-1 \). So, \( \sqrt{-49}=i\sqrt{49} \). Since \( \sqrt{49} = 7 \), we have \( \sqrt{-49}=7i \).

Step2: Simplify the expression \(-3\sqrt{-49}\)

Substitute \( \sqrt{-49}=7i \) into the expression: \(-3\times7i=-21i\).

Step3: Find the square root of \(-21i\)

Let \( z = x + yi \) be a complex number such that \( z^2=(x + yi)^2=-21i \). Expanding \( (x + yi)^2 \), we get \( x^2 - y^2 + 2xyi=-21i \). This gives us a system of equations:

1. \( x^2 - y^2 = 0 \) (real part)
2. \( 2xy=-21 \) (imaginary part)

From the first equation, \( x^2 = y^2 \), so \( x = y \) or \( x=-y \).

Case 1: \( x = y \)
Substitute into the second equation: \( 2x\times x=-21 \Rightarrow 2x^2=-21 \), which has no real solutions.

Case 2: \( x=-y \)
Substitute into the second equation: \( 2x\times(-x)=-21 \Rightarrow -2x^2=-21 \Rightarrow x^2=\frac{21}{2} \Rightarrow x=\pm\sqrt{\frac{21}{2}}=\pm\frac{\sqrt{42}}{2} \)

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 \( -\frac{\sqrt{42}}{2}+\frac{\sqrt{42}}{2}i \), which can be written as \( \pm\frac{\sqrt{42}}{2}(1 - i) \)

Answer:

\(\pm\frac{\sqrt{42}}{2}(1 - i)\) (or equivalently \(\pm\frac{\sqrt{42}}{2}-\frac{\sqrt{42}}{2}i\) and \(\pm\frac{\sqrt{42}}{2}+\frac{\sqrt{42}}{2}i\))