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Problem 499

AMC 12 late, AIME early

Algebra Difficulty 4.9 Find the answer

1.59 Given the simplest radical form $a \sqrt{2 a+b}$ and $\sqrt[a-b]{7}$ are like radicals, then the values of $a, b$ that satisfy the condition are
(A) do not exist.
(B) there is one set.
(C) there are two sets.
(D) there are more than two sets.
(China Junior High School Mathematics League, 1989)

Your answer

Official solution

[Solution] From the given conditions and the properties of similar radicals, we have
$\left\{\begin{array}{l} 2 a+b=7, \\ a-b=2 \text{. } \end{array}\right.$
Solving this, we get the unique solution:
$\left\{\begin{array}{l} a=3, \\ b=1 . \end{array}\right.$

Therefore, the correct choice is $(B)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.