Question

Use the Square Root Method to find the real solutions of ${(x+4)}^2=9$.

Short answer

Real solution of ${(x+4)}^2=9$is x = -1 and$x=-7$.

Step-by-step solution

Step 1. Given information.

We have given equation is,

${(x+4)}^2=9$

Step 2. Concept used.

If $a^2=bthena=\sqrt{b}anda=-\sqrt{b}$.

Step 3. Explanation.

We have given,

${(x+4)}^2=9$

Using square root method,

localid="1646649530505" $(x+4)=\sqrt{9}and(x+4)=-\sqrt{9}$

Solving both equations for x,

$$\begin{gathered} (x+4)=3and(x+4)=-3 \\ \Rightarrow x=3-4andx=-3-4 \\ \Rightarrow x=-1andx=-7 \end{gathered}$$

Step 4. Conclusion.

Hence, the real solution for ${(x+4)}^2=9$is

$x=-1$ and $x=-7$