High School

Solve [tex] 4|x+6| + 8 = 28 [/tex]

A. [tex] x = 1 [/tex] and [tex] x = -1 [/tex]

B. [tex] x = 1 [/tex] and [tex] x = -11 [/tex]

C. [tex] x = -1 [/tex] and [tex] x = 11 [/tex]

D. [tex] x = -1 [/tex] and [tex] x = -11 [/tex]

Answer :

To solve the equation [tex]\( 4|x+6| + 8 = 28 \)[/tex], let's follow these steps:

1. Isolate the absolute value expression:
[tex]\[
4|x+6| + 8 = 28
\][/tex]
Subtract 8 from both sides to isolate the term with the absolute value:
[tex]\[
4|x+6| = 20
\][/tex]

2. Divide by 4:
[tex]\[
|x+6| = 5
\][/tex]

3. Solve the absolute value equation:
The equation [tex]\( |x+6| = 5 \)[/tex] can be split into two separate equations:
[tex]\[
x + 6 = 5 \quad \text{and} \quad x + 6 = -5
\][/tex]

4. Solve these two equations:
[tex]\[
\begin{cases}
x + 6 = 5 \\
x + 6 = -5
\end{cases}
\][/tex]
- For [tex]\( x + 6 = 5 \)[/tex]:
[tex]\[
x = 5 - 6
\][/tex]
[tex]\[
x = -1
\][/tex]
- For [tex]\( x + 6 = -5 \)[/tex]:
[tex]\[
x = -5 - 6
\][/tex]
[tex]\[
x = -11
\][/tex]

5. Combine the solutions:
The solutions are [tex]\( x = -1 \)[/tex] and [tex]\( x = -11 \)[/tex].

So, the correct answer is:
[tex]\[
\boxed{D. \, x = -1 \, \text{and} \, x = -11}
\][/tex]