Problem 14 · AMC 8 Stretch
Core
Algebra & Patterns
intelligent-guessing-and-testingdetermining-necessary-conditions
Solve part (b): \(\sqrt{x - 2} = \sqrt{-1 - x}\). Square both sides to find a candidate, then check whether anything under a square root comes out negative. If there is a real solution give it; if not, answer 'none'.
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Answer: No real solution
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Hint 1 of 4
To get rid of the square-root signs, square both sides. But warning: squaring can create fake ('extraneous') answers, so you MUST check at the end.
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Hint 2 of 4
Squaring gives \(x - 2 = -1 - x\). Solve it, then check the insides of both square roots at your answer.
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Hint 3 of 4
You can't take the square root of a negative number and get a real number. If your candidate makes an inside negative, it's not a real solution.
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Approach: Square to get a candidate, then test the domain (necessary conditions)
- Square both sides: \(x - 2 = -1 - x\). Then \(2x = 1\), so the candidate is \(x = \frac{1}{2}\).
- Check the insides at \(x = \frac{1}{2}\): \(x - 2 = -\frac{3}{2}\) (negative) and \(-1 - x = -\frac{3}{2}\) (negative).
- Both square roots would be of negative numbers, which are not real. So \(x = \frac{1}{2}\) is a fake answer created by squaring.
- Therefore there is no real solution. (The companion equation (a) \(\sqrt{1-x} = \sqrt{x-1}\) does have a solution, \(x = 1\); the lesson is that squaring only gives a candidate — the check confirms it.)
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