To solve the quadratic equation [tex]\(4(x+2)^2 = 36\)[/tex], we will go through it step-by-step:
1. Divide both sides by 4: This simplifies the equation and makes it easier to solve.
[tex]\[
\frac{4(x+2)^2}{4} = \frac{36}{4} \implies (x+2)^2 = 9
\][/tex]
2. Take the square root of both sides: This will help to eliminate the squared term.
[tex]\[
\sqrt{(x+2)^2} = \sqrt{9} \implies x+2 = \pm 3
\][/tex]
Here, we consider both the positive and negative roots because taking the square root of a number results in two possible values.
3. Solve the two equations: We now have two separate equations to solve for [tex]\( x \)[/tex]:
[tex]\[
x+2 = 3 \quad \text{and} \quad x+2 = -3
\][/tex]
4. Solve for [tex]\( x \)[/tex] in each equation:
[tex]\[
x+2 = 3 \implies x = 3 - 2 \implies x = 1
\][/tex]
[tex]\[
x+2 = -3 \implies x = -3 - 2 \implies x = -5
\][/tex]
Therefore, the solutions to the equation [tex]\(4(x+2)^2 = 36\)[/tex] are [tex]\( x = -5 \)[/tex] and [tex]\( x = 1 \)[/tex].
Among the given choices:
- [tex]\( x = -11 \)[/tex] and [tex]\( x = 7 \)[/tex]
- [tex]\( x = -7 \)[/tex] and [tex]\( x = 11 \)[/tex]
- [tex]\( x = 5 \)[/tex] and [tex]\( x = 1 \)[/tex]
- [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex]
The correct answer is:
[tex]\[ \boxed{x = -5 \text{ and } x = 1} \][/tex]
