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Solve [tex]\( 4|x+6|=32 \)[/tex].

A. [tex]\( x=-14 \)[/tex] and [tex]\( x=-2 \)[/tex]

B. [tex]\( x=14 \)[/tex] and [tex]\( x=-2 \)[/tex]

C. [tex]\( x=-14 \)[/tex] and [tex]\( x=2 \)[/tex]

D. [tex]\( x=14 \)[/tex] and [tex]\( x=-14 \)[/tex]

Sagot :

GinnyAnswer
Let's solve the equation [tex]\( 4|x+6| = 32 \)[/tex] step-by-step:

1. Isolate the absolute value expression:
[tex]\[ 4|x+6| = 32 \][/tex]
Divide both sides by 4.
[tex]\[ |x+6| = \frac{32}{4} \][/tex]
[tex]\[ |x+6| = 8 \][/tex]

2. Set up the two cases for the absolute value equation:
The equation [tex]\( |x+6| = 8 \)[/tex] translates to two linear equations:
- Case 1: [tex]\( x + 6 = 8 \)[/tex]
- Case 2: [tex]\( x + 6 = -8 \)[/tex]

3. Solve for [tex]\( x \)[/tex] in each case:

- For Case 1:
[tex]\[ x + 6 = 8 \][/tex]
Subtract 6 from both sides:
[tex]\[ x = 8 - 6 \][/tex]
[tex]\[ x = 2 \][/tex]

- For Case 2:
[tex]\[ x + 6 = -8 \][/tex]
Subtract 6 from both sides:
[tex]\[ x = -8 - 6 \][/tex]
[tex]\[ x = -14 \][/tex]

4. Conclude the solutions:
The solutions to the equation [tex]\( 4|x+6| = 32 \)[/tex] are:
[tex]\[ x = -14 \quad \text{and} \quad x = 2 \][/tex]

Hence, the correct answer is:
[tex]\[ \boxed{C. \; x = -14 \text{ and } x = 2} \][/tex]
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