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solve $|6x + 36| = 12x$. identify the solution and an extraneous soluti…

Question

solve $|6x + 36| = 12x$. identify the solution and an extraneous solution.
a. solution: $x = -2$; extraneous solution: $x = 6$
b. solution: $x = 2$; extraneous solution: $x = -6$
c. solution: $x = 6$; extraneous solution: $x = -2$
d. solution: $x = 6$; extraneous solution: $x = -6$

Explanation:

Step1: Set up two cases for absolute value

Case 1: $6x + 36 = 12x$
Case 2: $6x + 36 = -12x$

Step2: Solve Case 1

Subtract $6x$ from both sides:
$36 = 6x$
Divide by 6: $x = 6$

Step3: Solve Case 2

Add $12x$ to both sides:
$18x + 36 = 0$
Subtract 36: $18x = -36$
Divide by 18: $x = -2$

Step4: Verify solutions

Check $x=6$: $|6(6)+36|=|36+36|=72$, $12(6)=72$. Valid.
Check $x=-2$: $|6(-2)+36|=|-12+36|=24$, $12(-2)=-24$. $24
eq -24$. Invalid.

Answer:

C. Solution: $x = 6$;
extraneous solution: $x = -2$