16Hit1600
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?

  1. A. Exactly oneCorrect
  2. B. Exactly two
  3. C. None
  4. 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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