To determine if \(^4 \sqrt{x} > ^3 \sqrt{x}\) given the statements, we need to analyze each statement individually.
- Analyzing Statement 1: \(0.5 < x < 2\)
- The inequality compares the fourth root and cube root of a positive number \(x\).
- To explore when \(^4 \sqrt{x} > ^3 \sqrt{x}\), consider the expressions:
\(x^{\frac{1}{4}} > x^{\frac{1}{3}}\) - Simplifying the inequality, we get \(x^{\frac{3}{12}} > x^{\frac{4}{12}}\), which implies \(x^3 > x^4\).
- Subtracting \(x^3\) from both sides gives: \(0 > x^4 - x^3\) or \(0 > x^3(x - 1)\).
- This indicates \(x^3 \gt 0\) and \((x-1) < 0\), which simplifies to \(x < 1\).
- Since \(0.5 < x < 2\) includes \(x > 1\), statement 1 is not conclusive. We cannot determine the outcome for all values in the range.
- Analyzing Statement 2: \(12x^2 - 7x + 1 = 0\)
- This is a quadratic equation. To find possible values for \(x\), use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 12, b = -7, c = 1\).
- The discriminant is \(b^2 - 4ac = 49 - 48 = 1\) (positive, implying two distinct real roots).
- Applying the quadratic formula, we find:
\(x = \frac{7 \pm 1}{24}\), giving solutions \(x = \frac{8}{24} = \frac{1}{3}\) and \(x = \frac{6}{24} = \frac{1}{4}\). - Both solutions, \(x = \frac{1}{3}\) and \(x = \frac{1}{4}\), satisfy \(x < 1\).
- Thus, for both values, \(^4 \sqrt{x} > ^3 \sqrt{x}\) holds true since \(x < 1\).
Conclusion: Statement 2 alone is sufficient to determine that \(^4 \sqrt{x} > ^3 \sqrt{x}\), while Statement 1 is inconclusive. Therefore, the correct answer is: statement (2) alone is sufficient to answer the question.