Question

The solution of the equation \( y^{\frac{2}{3}} - 2y^{\frac{1}{3}} = 15 \) is:

• A. 25, 27
• B. 27, -125
• C. 25, -27
• D. 125, -27

Hint:

When solving equations with fractional exponents, first substitute a variable for the fractional power, solve the quadratic equation, and then substitute back.

Answer 1

The Correct Option is C

Let \( z = y^{\frac{1}{3}} \), then \( z^2 = y^{\frac{2}{3}} \), and the equation becomes: \[ z^2 - 2z = 15 \] Rearranging: \[ z^2 - 2z - 15 = 0 \] Factoring the quadratic equation: \[ (z - 5)(z + 3) = 0 \] Thus, \( z = 5 \) or \( z = -3 \). Since \( z = y^{\frac{1}{3}} \), we have: \[ y^{\frac{1}{3}} = 5 \quad \Rightarrow \quad y = 25 \] \[ y^{\frac{1}{3}} = -3 \quad \Rightarrow \quad y = -27 \] Therefore, the solutions are \( y = 25 \) and \( y = -27 \).