Q. 350 - Intermediate Algebra

Question

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

(b) Explain why one of the solution that was found

was not actually a solution to the equation.

Step-by-Step Solution

Answer

The value of $r=5.$

Step 1. Given information

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

Part (a) Step 1. Solution of the equation

$$\begin{gathered} \sqrt{r+4}-r+2=0 \\ \sqrt{r+4}=r-2 \end{gathered}$$

Squaring Both sides

$$\begin{gathered} \left(r+4\right)={\left(r-2\right)}^2 \\ r+4=r^2-4r+4 \\ {r}^2-4r+4-r-4=0 \\ {r}^2-5r=0 \\ r\left(r-5\right)=0 \end{gathered}$$

Part (a) Step 2. Solve by zero property

$$\begin{gathered} r=0,r-5 \\ r=0,r=5 \end{gathered}$$

Part (b) Step 1. One solution was not found solution

0 is not a solution of the equation because it turns 

$$\begin{gathered} \sqrt{0+4}-0+2=0 \\ 2+2=0 \\ 4=0 \end{gathered}$$

which is not possible.