Asked by David
Which of the following is an extraneous solution of StartRoot 4 x + 41 EndRoot = x + 5?
x = –8
x = –2
All Answers 1
Answered by
GPT-5 mini
AI
Solve sqrt(4x + 41) = x + 5. Domain: 4x + 41 ≥ 0 ⇒ x ≥ -10.25.
Square both sides: 4x + 41 = (x + 5)^2 = x^2 + 10x + 25
⇒ 0 = x^2 + 6x - 16 = (x - 2)(x + 8)
⇒ x = 2 or x = -8.
Check in the original:
- x = 2: sqrt(49) = 7 = 2 + 5 → valid.
- x = -8: sqrt(9) = 3 ≠ -8 + 5 = -3 → extraneous.
Thus x = –8 is the extraneous solution. (x = –2 is not a solution at all.)
Square both sides: 4x + 41 = (x + 5)^2 = x^2 + 10x + 25
⇒ 0 = x^2 + 6x - 16 = (x - 2)(x + 8)
⇒ x = 2 or x = -8.
Check in the original:
- x = 2: sqrt(49) = 7 = 2 + 5 → valid.
- x = -8: sqrt(9) = 3 ≠ -8 + 5 = -3 → extraneous.
Thus x = –8 is the extraneous solution. (x = –2 is not a solution at all.)
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