High School

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 :

Sure! Let's solve the equation [tex]\(4|x+5|=28\)[/tex] step by step.

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

2. Consider the definition of absolute value: The absolute value [tex]\(|x + 5|\)[/tex] can be either positive 7 or negative 7. This gives us two scenarios:
[tex]\[
x + 5 = 7
\][/tex]
[tex]\[
x + 5 = -7
\][/tex]

3. Solve the first scenario [tex]\(x + 5 = 7\)[/tex]:
[tex]\[
x + 5 = 7
\][/tex]
Subtract 5 from both sides:
[tex]\[
x = 7 - 5
\][/tex]
[tex]\[
x = 2
\][/tex]

4. Solve the second scenario [tex]\(x + 5 = -7\)[/tex]:
[tex]\[
x + 5 = -7
\][/tex]
Subtract 5 from both sides:
[tex]\[
x = -7 - 5
\][/tex]
[tex]\[
x = -12
\][/tex]

So, the solutions are [tex]\(x = 2\)[/tex] and [tex]\(x = -12\)[/tex]. Therefore, the correct option is:
[tex]\[
\boxed{\text{C. } x = -12 \text{ and } x = 2}
\][/tex]

Other Questions