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12 Jan 2026, 07:38
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If \(x\) is a non-negative number and \(x ≠ 3\sqrt{x} + 4\), which of the following must be true?
I. \((x - 1)(x - 16) ≠ 0\)
II. \(|17 - x| ≠ 1\)
III. \(x\) is not a square of an integer
A. I only
B. II only
C. III only
D. I, and II only
E. None of the above
I. \((x - 1)(x - 16) ≠ 0\)
II. \(|17 - x| ≠ 1\)
III. \(x\) is not a square of an integer
A. I only
B. II only
C. III only
D. I, and II only
E. None of the above
12 Jan 2026, 07:38
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Official Solution:
If \(x\) is a non-negative number and \(x ≠ 3\sqrt{x} + 4\), which of the following must be true?
I. \((x - 1)(x - 16) ≠ 0\)
II. \(|17 - x| ≠ 1\)
III. \(x\) is not a square of an integer
A. I only
B. II only
C. III only
D. I, and II only
E. None of the above
Let's solve \(x = 3\sqrt{x} + 4\).
Let \(\sqrt{x} = t\), so \(x = t^2\).
Substitute:
\(t^2 = 3t + 4\)
\(t^2 - 3t - 4 = 0\)
Factor:
\((t - 4)(t + 1) = 0\)
So \(t = 4\) or \(t = -1\).
But \(t = \sqrt{x}\) cannot be negative, so only \(t = 4\) is valid.
Thus \(\sqrt{x} = 4\), so \(x = 16\).
Therefore \(x \ne 3\sqrt{x} + 4\) means \(x \ne 16\). That is the only restriction (along with \(x\) being a non-negative number).
Now test each statement to see whether it must be true when \(x \ne 16\).
I. \((x - 1)(x - 16) \ne 0\)
This expression is nonzero unless \(x = 1\) or \(x = 16\). We only know \(x \ne 16\), but \(x\) could be 1, which makes the expression 0. So Statement I is not guaranteed. Not always true.
II. \(|17 - x| \ne 1\)
\(|17 - x| = 1\) corresponds to \(x = 16\) or \(x = 18\). We only know \(x \ne 16\), but \(x\) could be 18, which would make \(|17 - x| = 1\). So Statement II is not guaranteed. Not always true.
III. \(x\) is not a square of an integer
We only know \(x \ne 16\), but \(x\) could be 0, 1, 4, 9, 25, 36, etc. So Statement III is not guaranteed. Not always true.
Answer: E
If \(x\) is a non-negative number and \(x ≠ 3\sqrt{x} + 4\), which of the following must be true?
I. \((x - 1)(x - 16) ≠ 0\)
II. \(|17 - x| ≠ 1\)
III. \(x\) is not a square of an integer
A. I only
B. II only
C. III only
D. I, and II only
E. None of the above
Let's solve \(x = 3\sqrt{x} + 4\).
Let \(\sqrt{x} = t\), so \(x = t^2\).
Substitute:
\(t^2 = 3t + 4\)
\(t^2 - 3t - 4 = 0\)
Factor:
\((t - 4)(t + 1) = 0\)
So \(t = 4\) or \(t = -1\).
But \(t = \sqrt{x}\) cannot be negative, so only \(t = 4\) is valid.
Thus \(\sqrt{x} = 4\), so \(x = 16\).
Therefore \(x \ne 3\sqrt{x} + 4\) means \(x \ne 16\). That is the only restriction (along with \(x\) being a non-negative number).
Now test each statement to see whether it must be true when \(x \ne 16\).
I. \((x - 1)(x - 16) \ne 0\)
This expression is nonzero unless \(x = 1\) or \(x = 16\). We only know \(x \ne 16\), but \(x\) could be 1, which makes the expression 0. So Statement I is not guaranteed. Not always true.
II. \(|17 - x| \ne 1\)
\(|17 - x| = 1\) corresponds to \(x = 16\) or \(x = 18\). We only know \(x \ne 16\), but \(x\) could be 18, which would make \(|17 - x| = 1\). So Statement II is not guaranteed. Not always true.
III. \(x\) is not a square of an integer
We only know \(x \ne 16\), but \(x\) could be 0, 1, 4, 9, 25, 36, etc. So Statement III is not guaranteed. Not always true.
Answer: E
