Q 136E - Intermediate Algebra

Question

In the following exercises, simplify.

Part (a): ${\left(-49\right)}^{\frac{1}{2}}$

Part (b): $-{49}^{\frac{1}{2}}$

Part (c): ${\left(49\right)}^{-\frac{1}{2}}$

Step-by-Step Solution

Answer

Part (a): On simplifying, we know the solution is not real.

Part (b): On simplifying, we get $-7$.

Part (c): On simplifying, we get $\frac{1}{7}$.

Part (a) Step 1. Rewrite as a root.

Consider the given question,

${\left(-49\right)}^{\frac{1}{2}}$

Rewrite as a square root,

$=\sqrt[2]{\left(-49\right)}$

On simplifying the expression,

$=\sqrt[2]{{\left(-7\right)}^2}$

The solution is not real.

Part (b) Step 1. Rewrite as a root.

Consider the given question,

$-{49}^{\frac{1}{2}}$

The exponent only applies to the $49$.

Rewrite as a square root,

$=-\sqrt[2]{49}$

On simplifying the expression,

$$\begin{gathered} =-\sqrt[2]{7^2} \\ =-7 \end{gathered}$$

Part (c) Step 1. Rewrite as a root.

Consider the given question,

${\left(49\right)}^{-\frac{1}{2}}$

Rewrite using the property $a^{-n}=\frac{1}{a^n}$,

$=\frac{1}{{49}^{{\displaystyle \frac{1}{2}}}}$

Rewrite as a square root,

$=\frac{1}{\sqrt[2]{49}}$

On simplifying the expression,

$$\begin{gathered} =\frac{1}{\sqrt[2]{7^2}} \\ =\frac{1}{7} \end{gathered}$$