College

Solve [tex]$4|x+5|=28$[/tex].

A. [tex]$x=12$[/tex] and [tex]$x=-2$[/tex]
B. [tex]$x=-12$[/tex] and [tex]$x=-2$[/tex]
C. [tex]$x=12$[/tex] and [tex]$x=-12$[/tex]
D. [tex]$x=-12$[/tex] and [tex]$x=2$[/tex]

Answer :

To solve the equation [tex]\(4|x+5|=28\)[/tex], follow these steps:

1. Isolate the absolute value:
Divide both sides of the equation by 4:
[tex]\[
|x+5| = \frac{28}{4} = 7
\][/tex]

2. Solve for the two cases of the absolute value:

- Case 1: [tex]\(x + 5 = 7\)[/tex]
- Subtract 5 from both sides:
[tex]\[
x = 7 - 5 = 2
\][/tex]

- Case 2: [tex]\(x + 5 = -7\)[/tex]
- Subtract 5 from both sides:
[tex]\[
x = -7 - 5 = -12
\][/tex]

Thus, the solutions to the equation are [tex]\(x = 2\)[/tex] and [tex]\(x = -12\)[/tex].

Comparing these solutions with the given options:
- Option A: [tex]\(x=12\)[/tex] and [tex]\(x=-2\)[/tex]
- Option B: [tex]\(x=-12\)[/tex] and [tex]\(x=2\)[/tex]
- Option C: [tex]\(x=12\)[/tex] and [tex]\(x=-12\)[/tex]
- Option D: [tex]\(x=-12\)[/tex] and [tex]\(x=2\)[/tex]

The correct choice is Option B: [tex]\(x=-12\)[/tex] and [tex]\(x=2\)[/tex].