Question

If \( \alpha, \beta \) are the real roots of the equation \( 12x^\frac{1}{3} - 25x^\frac{1}{6} + 12 = 0 \), if \( \alpha>\beta \), then \( 6\sqrt{\frac{\alpha}{\beta}} = \):

• A. \( \frac{3}{2} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{9}{8} \)
• D. \( \frac{16}{9} \) \bigskip

Hint:

Substitute to simplify higher degree equations into quadratic forms, then use the quadratic formula to solve.

Answer 1

The Correct Option is D

We are given the equation: \[ 12x^{\frac{1}{3}} - 25x^{\frac{1}{6}} + 12 = 0 \] Step 1: Substituting a New Variable Let: \[ y = x^{\frac{1}{6}} \] Then: \[ x^{\frac{1}{3}} = y^2 \] Rewriting the equation in terms of \( y \): \[ 12y^2 - 25y + 12 = 0 \] Step 2: Solve for \( y \) Using the quadratic formula: \[ y = \frac{-(-25) \pm \sqrt{(-25)^2 - 4(12)(12)}}{2(12)} \] \[ = \frac{25 \pm \sqrt{625 - 576}}{24} \] \[ = \frac{25 \pm \sqrt{49}}{24} \] \[ = \frac{25 \pm 7}{24} \] \[ y = \frac{32}{24} = \frac{4}{3}, \quad y = \frac{18}{24} = \frac{3}{4} \] Step 3: Compute \( \frac{\alpha}{\beta} \) Since \( x = y^6 \), we find: \[ \alpha = \left(\frac{4}{3}\right)^6, \quad \beta = \left(\frac{3}{4}\right)^6 \] \[ \frac{\alpha}{\beta} = \left(\frac{4}{3}\right)^{12} \] Step 4: Compute \( 6\sqrt{\frac{\alpha}{\beta}} \) \[ 6\sqrt{\frac{\alpha}{\beta}} = 6 \times \left(\frac{4}{3}\right)^6 \] \[ = 6 \times \frac{4096}{729} \] \[ = \frac{24576}{729} = \frac{16}{9} \] Final Answer: \(\boxed{\frac{16}{9}}\) \bigskip