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Solve [tex]|6x + 18| = 12x[/tex]. Identify the solution and an extraneous solution.

A. Solution: [tex]x = 3[/tex]; extraneous solution: [tex]x = -3[/tex]
B. Solution: [tex]x = 3[/tex]; extraneous solution: [tex]x = -1[/tex]
C. Solution: [tex]x = 1[/tex]; extraneous solution: [tex]x = -3[/tex]
D. Solution: [tex]x = -1[/tex]; extraneous solution: [tex]x = 3[/tex]


Sagot :

Let's solve the equation [tex]\(|6x + 18| = 12x\)[/tex] and determine which solutions are valid and which are extraneous. Here's a step-by-step breakdown:

1. Break Down the Absolute Value:
The equation [tex]\(|6x + 18| = 12x\)[/tex] can be split into two cases because the absolute value of [tex]\(6x + 18\)[/tex] depends on whether [tex]\(6x + 18\)[/tex] is non-negative or negative.

- Case 1: [tex]\(6x + 18 \geq 0\)[/tex]
In this case, [tex]\(|6x + 18| = 6x + 18\)[/tex].
Therefore, the equation becomes:
[tex]\[ 6x + 18 = 12x \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 18 = 6x \][/tex]
[tex]\[ x = 3 \][/tex]

- Case 2: [tex]\(6x + 18 < 0\)[/tex]
In this case, [tex]\(|6x + 18| = -(6x + 18) = -6x - 18\)[/tex].
Therefore, the equation becomes:
[tex]\[ -6x - 18 = 12x \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ -18 = 18x \][/tex]
[tex]\[ x = -1 \][/tex]

2. Check for Validity:
Now, we need to check whether these solutions are valid by substituting them back into the original absolute value equation.

- For [tex]\(x = 3\)[/tex]:
[tex]\[ |6(3) + 18| = 12(3) \][/tex]
[tex]\[ |18 + 18| = 36 \][/tex]
[tex]\[ |36| = 36 \quad \text{(True)} \][/tex]

- For [tex]\(x = -1\)[/tex]:
[tex]\[ |6(-1) + 18| = 12(-1) \][/tex]
[tex]\[ |-6 + 18| = -12 \][/tex]
[tex]\[ |12| = -12 \quad \text{(False, because absolute values cannot be negative)} \][/tex]

Therefore, the correct solution to the equation is [tex]\(x = 3\)[/tex] and the extraneous solution is [tex]\(x = -1\)[/tex].

Thus, the answer is:
B. Solution: [tex]\(x=3\)[/tex]; extraneous solution: [tex]\(x=-1\)[/tex]
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