College

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

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

Answer :

To solve the equation [tex]\(4|x+5|=28\)[/tex], we begin by isolating the absolute value expression.

1. Divide both sides by 4:
[tex]\[
|x+5| = \frac{28}{4}
\][/tex]
[tex]\[
|x+5| = 7
\][/tex]

2. Split the absolute value equation into two separate cases. This is because the absolute value of a number, [tex]\(a\)[/tex], can be both [tex]\(a\)[/tex] and [tex]\(-a\)[/tex].

- Case 1:
[tex]\[
x+5 = 7
\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[
x = 7 - 5
\][/tex]
[tex]\[
x = 2
\][/tex]

- Case 2:
[tex]\[
x+5 = -7
\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[
x = -7 - 5
\][/tex]
[tex]\[
x = -12
\][/tex]

3. The solutions for [tex]\(x\)[/tex] are [tex]\( x = 2 \)[/tex] and [tex]\( x = -12 \)[/tex].

So, the correct answer is C. [tex]\(x=-12\)[/tex] and [tex]\(x=2\)[/tex].