Question

use the imaginary number i to rewrite the expression below all radicals.
-12 - √(-20)
-12 - i√20

Explanation:

Step1: Recall the definition of \(i\)

Recall that \(i = \sqrt{-1}\), so \(\sqrt{-a}=\sqrt{-1\times a}=i\sqrt{a}\) for \(a\geq0\).

Step2: Apply the definition to \(\sqrt{-20}\)

For \(\sqrt{-20}\), we can rewrite it as \(\sqrt{-1\times20}\). Using the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (for \(a = - 1\) and \(b = 20\) where \(a\) is a negative real number and \(b\) is a non - negative real number), we have \(\sqrt{-20}=\sqrt{-1}\times\sqrt{20}\). Since \(\sqrt{-1}=i\), then \(\sqrt{-20}=i\sqrt{20}\). But in the original expression, we have \(-\sqrt{-20}\), so \(-\sqrt{-20}=-i\sqrt{20}\).

Step3: Rewrite the original expression

The original expression is \(-12-\sqrt{-20}\). Substituting \(-\sqrt{-20}=-i\sqrt{20}\) into the expression, we get \(-12 - i\sqrt{20}\). We can also simplify \(\sqrt{20}\) as \(2\sqrt{5}\), so the expression can be further written as \(-12-2i\sqrt{5}\), but the form \(-12 - i\sqrt{20}\) is also correct.

Answer:

\(-12 - i\sqrt{20}\) (or \(-12-2i\sqrt{5}\))