SAT Math: Radical Equations and Extraneous Solutions Practice Questions
Isolate square roots, square both sides carefully, and test every candidate against the original equation. Try each original question before checking its worked explanation. This focused practice is free and requires no account.
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Answer the questions below — you get instant feedback and a full explanation for each.
1. What value of x satisfies sqrt(2x + 5) = 5, where sqrt denotes the principal square root?
Explanation. Square both sides to get 2x + 5 = 25, so x = 10. Checking: sqrt(25) = 5. The principal square root is nonnegative.
2. What is the only real solution of sqrt(x + 6) = x?
Explanation. The right side must be nonnegative. Squaring gives x + 6 = x^2, or (x - 3)(x + 2) = 0. Candidates are 3 and -2; only 3 satisfies the original equation, since sqrt(4) is 2, not -2.
3. What value of x satisfies sqrt(3x + 1) = 4?
Explanation. Squaring gives 3x + 1 = 16, so 3x = 15 and x = 5. Substitution produces sqrt(16) = 4.
4. What value of x satisfies sqrt(x - 1) + 2 = 6?
Explanation. First isolate the radical: sqrt(x - 1) = 4. Squaring gives x - 1 = 16, so x = 17. The original left side is 4 + 2 = 6.
5. How many real solutions does sqrt(x + 4) = -2 have?
Explanation. A principal square root is always nonnegative, so it cannot equal -2. Squaring would produce x = 0, but checking gives 2 = -2, which is false.
6. What is the only real solution of sqrt(2x + 3) = x?
Explanation. Since the square root equals x, x must be nonnegative. Squaring gives x^2 - 2x - 3 = 0, or (x - 3)(x + 1) = 0. Only x = 3 is nonnegative and passes substitution: sqrt(9) = 3.
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FAQ
Why can squaring create an extraneous solution?
Opposite numbers have equal squares. Squaring loses the sign information in the original equation, so each candidate must be substituted back.
What should I do before squaring?
Isolate the radical and consider the signs and domain. A nonnegative square root cannot equal a negative expression at a valid solution.