Question

(a) Solve the equation $\sqrt{\mathbf{r}\mathbf{+}\mathbf{4}}-\mathit{r}\mathbf{+}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{0}$

(b) Explain why one of the “solutions” that was found was not actually a solution to the equation.

Short answer

Part (a) : r = 5,0

Part (b) : r = 0 is an extraneous solution

Step-by-step solution

Step 1. Given Information:

$\sqrt{\mathbf{r}\mathbf{+}\mathbf{4}}-\mathit{r}\mathbf{+}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{0}$

Step 2. Finding solutions for part (a) :

$$\begin{gathered} \sqrt{\mathbf{r}\mathbf{+}\mathbf{4}}-\mathit{r}\mathbf{+}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{0} \\ \sqrt{\mathbf{r}\mathbf{+}\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{}\mathit{r}\mathbf{-}\mathbf{2}\mathbf{} \\ \mathit{r}\mathbf{+}\mathbf{4}\mathbf{=}{\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{2}} \\ \mathit{r}\mathbf{+}\cancel{\mathbf{4}}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{+}\cancel{\mathbf{4}}\mathbf{-}\mathbf{4}\mathit{r} \\ \mathbf{0}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathit{r} \\ \mathbf{0}\mathbf{=}\mathit{r}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{5}\mathbf{)} \\ \mathit{r}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{;}\mathbf{}\mathit{r}\mathbf{=}\mathbf{5}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathit{U}\mathit{s}\mathit{e}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{z}\mathit{e}\mathit{r}\mathit{o}\mathbf{}\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathbf{}\mathit{p}\mathit{r}\mathit{o}\mathit{p}\mathit{e}\mathit{r}\mathit{t}\mathit{y}\mathbf{)} \end{gathered}$$

Step 3. Check the solution for part(b) :

$$\begin{gathered} \mathit{r}\mathbf{=}\mathbf{5} \\ \sqrt{\mathbf{5}\mathbf{+}\mathbf{4}}\mathbf{}\mathbf{-}\mathbf{5}\mathbf{+}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{0} \\ \sqrt{\mathbf{9}}\mathbf{}\mathbf{-}\mathbf{3}\mathbf{}\mathbf{=}\mathbf{0} \\ \mathbf{3}\mathbf{-}\mathbf{3}\mathbf{=}\mathbf{0} \\ \mathbf{0}\mathbf{=}\mathbf{0} \\ \mathit{s}\mathit{o}\mathbf{,}\mathbf{}\mathbf{5}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{o}\mathit{n}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{i}\mathit{t}\mathit{s}\mathbf{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{.} \\ \mathit{r}\mathbf{=}\mathbf{0} \\ \sqrt{\mathbf{0}\mathbf{+}\mathbf{4}}\mathbf{}\mathbf{-}\mathbf{0}\mathbf{+}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{0} \\ \sqrt{\mathbf{4}}\mathbf{}\mathbf{+}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{0} \\ \mathbf{2}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{0} \\ \mathbf{4}\mathbf{\ne }\mathbf{0} \\ \mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{a}\mathit{n}\mathbf{}\mathit{e}\mathit{x}\mathit{t}\mathit{r}\mathit{a}\mathit{n}\mathit{e}\mathit{o}\mathit{u}\mathit{s}\mathbf{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{.} \end{gathered}$$