Answer:
\(x=-5\text{ or }x=1\).
Step-by-step explanation:
We have been given a quadratic equation \(4(x+2)^2=36\). We are asked to find the solutions for our given equation.
First of all, we will divide both sides of our given equation by 4 as shown below:
\(\frac{4(x+2)^2}{4}=\frac{36}{4}\)
\((x+2)^2=9\)
Now, we will take square root of both sides of our equation.
\(\sqrt{(x+2)^2}=\pm \sqrt{9}\)
\(\sqrt{(x+2)^2}=\pm \sqrt{3^2}\)
Using radical rule \(\sqrt[n]{a^n}=a\), we will get:
\(x+2=\pm 3\)
Upon subtracting 2 from both sides of our given equation, we will get:
\(x+2-2=-2\pm 3\)
\(x=-2\pm 3\)
Now, we will write two equivalent equations to our equation as:
\(x=-2-3\text{ or }x=-2+3\)
\(x=-5\text{ or }x=1\)
Therefore, the solutions for our given quadratic equation are \(x=-5\text{ or }x=1\).
