Advanced MathVery hardCounting real solutions of a radical equation
Advanced Math practice question
How many real solutions does the equation √(x + 7) = x − 5 have?
- A. Exactly oneCorrect
- B. Exactly two
- C. None
- D. Infinitely many
Answer: A. Exactly one
Squaring gives x + 7 = x² − 10x + 25, so x² − 11x + 18 = (x − 2)(x − 9) = 0. Check x = 2: √9 = 3 but 2 − 5 = −3, rejected. Check x = 9: √16 = 4 = 9 − 5 ✓. Exactly one solution.
Why the other answers are wrong
- B. Exactly two
- Squaring produces two candidates, but x = 2 makes the right side negative and fails the original equation.
- C. None
- x = 9 does satisfy the equation (4 = 4).
- D. Infinitely many
- A radical equation of this kind has at most two candidates.
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