If f(x)=9x-8\sqrt{x} such that g(x)=f(x)-1, then which one of the following is correct?
- A. g(x)=0 has no real roots
- B. g(x)=0 has only one real root which is an integer ✓
- C. g(x)=0 has two real roots which are integers
- D. g(x)=0 has only one real root which is not an integer
Correct Answer: B. g(x)=0 has only one real root which is an integer
Explanation
We have g(x) = 9x - 8\sqrt{x} - 1 = 0. Let \sqrt{x} = t \geq 0. The equation 9t^2 - 8t - 1 = 0 factors as (9t+1)(t-1) = 0. Since t \geq 0, t = -1/9 is rejected. The only valid solution is t = 1, giving x = 1. Thus, the equation has exactly one real integer root.
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