Asked by deja
A student stated that there are two solutions to the equation √(3x + 7) = x - 1.
They are x = -1 and x = 6.
All Answers 1
Answered by
GPT-5 mini
AI
They're incorrect. Work through it:
1. Domain: √(3x+7) is defined for 3x+7 ≥ 0 ⇒ x ≥ −7/3, and the left side is nonnegative so the right side must be nonnegative: x−1 ≥ 0 ⇒ x ≥ 1. So any solution must satisfy x ≥ 1.
2. Square both sides: 3x+7 = (x−1)^2 = x^2 − 2x + 1. Rearranging gives x^2 − 5x − 6 = 0, whose roots are x = 6 and x = −1.
3. Check each in the original equation:
- x = 6: √(3·6+7) = √25 = 5 and 6−1 = 5 → valid.
- x = −1: √(3·(−1)+7) = √4 = 2 and −1−1 = −2 → not equal, so −1 is extraneous.
Only x = 6 satisfies the original equation. The extraneous root arose from squaring both sides (which can introduce solutions that don't satisfy the original sign constraints).
1. Domain: √(3x+7) is defined for 3x+7 ≥ 0 ⇒ x ≥ −7/3, and the left side is nonnegative so the right side must be nonnegative: x−1 ≥ 0 ⇒ x ≥ 1. So any solution must satisfy x ≥ 1.
2. Square both sides: 3x+7 = (x−1)^2 = x^2 − 2x + 1. Rearranging gives x^2 − 5x − 6 = 0, whose roots are x = 6 and x = −1.
3. Check each in the original equation:
- x = 6: √(3·6+7) = √25 = 5 and 6−1 = 5 → valid.
- x = −1: √(3·(−1)+7) = √4 = 2 and −1−1 = −2 → not equal, so −1 is extraneous.
Only x = 6 satisfies the original equation. The extraneous root arose from squaring both sides (which can introduce solutions that don't satisfy the original sign constraints).
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