Question

use the imaginary number i to rewrite the expression below as a complex number. simplify all radicals.
$-sqrt{-60}$

Explanation:

Step1: Recall the definition of \(i\)

We know that \(i = \sqrt{-1}\), so we can rewrite \(\sqrt{-60}\) as \(\sqrt{60\times(-1)}\).
Using the property of square roots \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (for \(a\geq0, b\geq0\), here we extend it to complex numbers), we have \(\sqrt{60\times(-1)}=\sqrt{60}\cdot\sqrt{-1}\).
Since \(\sqrt{-1} = i\), this becomes \(\sqrt{60}\cdot i\).

Step2: Simplify \(\sqrt{60}\)

We factor \(60\) into \(4\times15\), so \(\sqrt{60}=\sqrt{4\times15}\).
Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (for \(a\geq0, b\geq0\)), we get \(\sqrt{4\times15}=\sqrt{4}\cdot\sqrt{15}\).
Since \(\sqrt{4} = 2\), this simplifies to \(2\sqrt{15}\).

Step3: Combine with the negative sign and \(i\)

The original expression is \(-\sqrt{-60}\), and from Step 1 and Step 2, \(\sqrt{-60}=2\sqrt{15}i\), so \(-\sqrt{-60}=-2\sqrt{15}i\).

Answer:

\(-2\sqrt{15}i\)