Question

1. rewrite -7i as a radical.

a. \\(sqrt{-49}\\)
b. \\(sqrt{-7}\\)
c. \\(-sqrt{-7}\\)
d. \\(-sqrt{-49}\\)

Explanation:

Step1: Recall the definition of imaginary unit \(i\)

We know that \(i = \sqrt{-1}\), so we can rewrite \(-7i\) by expressing the coefficient in terms of a square root.
First, let's recall that for a real number \(a\), \(ai=a\sqrt{-1}\), and also \(\sqrt{-b}=\sqrt{b}\cdot\sqrt{-1}=i\sqrt{b}\) for \(b>0\). We want to express \(-7i\) as a radical. Let's start by manipulating the expression.

We know that \(i=\sqrt{-1}\), so \(- 7i=-7\sqrt{-1}\). Now, we can rewrite \(-7\) as \(-\sqrt{49}\) (since \(\sqrt{49} = 7\)). So we have:

\(-7i=-\sqrt{49}\cdot\sqrt{-1}\)

Step2: Use the property of radicals \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) (for \(a,b\) such that the product is defined in the complex or real number system, here \(a = 49\) and \(b=- 1\))

Using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\), we get:

\(-\sqrt{49}\cdot\sqrt{-1}=-\sqrt{49\times(- 1)}=-\sqrt{-49}\)

Answer:

d. \(-\sqrt{-49}\)