21. \( 2 x^{\frac{1}{2}}-x^{\frac{1}{4}}=0 \)

To solve the equation: \[ 2x^{\frac{1}{2}} - x^{\frac{1}{4}} = 0 \] let's follow these steps: ### Step 1: Substitute to Simplify Let \( y = x^{\frac{1}{4}} \). Then, \( y^2 = x^{\frac{2}{4}} = x^{\frac{1}{2}} \). Substituting these into the original equation: \[ 2y^2 - y = 0 \] ### Step 2: Factor the Equation Factor out \( y \): \[ y(2y - 1) = 0 \] This gives two possible solutions for \( y \): 1. \( y = 0 \) 2. \( 2y - 1 = 0 \Rightarrow y = \frac{1}{2} \) ### Step 3: Solve for \( x \) Recall that \( y = x^{\frac{1}{4}} \). Now, solve for \( x \) in each case: 1. **When \( y = 0 \):** \[ x^{\frac{1}{4}} = 0 \Rightarrow x = 0^4 = 0 \] 2. **When \( y = \frac{1}{2} \):** \[ x^{\frac{1}{4}} = \frac{1}{2} \Rightarrow x = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Step 4: Verify the Solutions It's always good practice to verify the solutions in the original equation. 1. **For \( x = 0 \):** \[ 2(0)^{\frac{1}{2}} - (0)^{\frac{1}{4}} = 0 - 0 = 0 \quad \text{✓} \] 2. **For \( x = \frac{1}{16} \):** \[ 2\left(\frac{1}{16}\right)^{\frac{1}{2}} - \left(\frac{1}{16}\right)^{\frac{1}{4}} = 2\left(\frac{1}{4}\right) - \left(\frac{1}{2}\right) = \frac{1}{2} - \frac{1}{2} = 0 \quad \text{✓} \] ### Conclusion The solutions to the equation \( 2x^{\frac{1}{2}} - x^{\frac{1}{4}} = 0 \) are: \[ x = 0 \quad \text{and} \quad x = \frac{1}{16} \]