Asked by knmk
What is the solution of StartRoot x + 12 EndRoot = x?
x = –3
x = 4
x = –3 or x = 4
no solution
All Answers 1
Answered by
GPT-5 mini
AI
Solve sqrt(x + 12) = x.
Domain: x + 12 ≥ 0 ⇒ x ≥ -12. Square both sides: x + 12 = x^2 ⇒ x^2 - x - 12 = 0 ⇒ (x - 4)(x + 3) = 0 ⇒ x = 4 or x = -3.
Check both in the original equation:
- x = 4: sqrt(4 + 12) = sqrt(16) = 4 ✔
- x = -3: sqrt(-3 + 12) = sqrt(9) = 3 ≠ -3 ✖
So the only solution is x = 4.
Domain: x + 12 ≥ 0 ⇒ x ≥ -12. Square both sides: x + 12 = x^2 ⇒ x^2 - x - 12 = 0 ⇒ (x - 4)(x + 3) = 0 ⇒ x = 4 or x = -3.
Check both in the original equation:
- x = 4: sqrt(4 + 12) = sqrt(16) = 4 ✔
- x = -3: sqrt(-3 + 12) = sqrt(9) = 3 ≠ -3 ✖
So the only solution is x = 4.
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