(1) \( x-3 x^{\frac{1}{2}}+2=0 \) (3) \( x^{\frac{1}{2}}+3 x^{\frac{1}{4}}-10=0 \)

Sure, let's solve each of the given equations step by step. --- ### **Problem (1):** \[ x - 3x^{\frac{1}{2}} + 2 = 0 \] #### **Solution:** 1. **Substitution:** Let \( y = \sqrt{x} \). Then, \( x = y^2 \). 2. **Rewrite the Equation:** Substitute \( x = y^2 \) into the original equation: \[ y^2 - 3y + 2 = 0 \] 3. **Solve the Quadratic Equation:** \[ y^2 - 3y + 2 = 0 \] Factor the quadratic: \[ (y - 1)(y - 2) = 0 \] So, \( y = 1 \) or \( y = 2 \). 4. **Back-Substitute to Find \( x \):** - If \( y = 1 \): \[ \sqrt{x} = 1 \implies x = 1^2 = 1 \] - If \( y = 2 \): \[ \sqrt{x} = 2 \implies x = 2^2 = 4 \] 5. **Solutions:** \[ x = 1 \quad \text{and} \quad x = 4 \] --- ### **Problem (3):** \[ x^{\frac{1}{2}} + 3x^{\frac{1}{4}} - 10 = 0 \] #### **Solution:** 1. **Substitution:** Let \( y = x^{\frac{1}{4}} \). Then, \( x^{\frac{1}{2}} = y^2 \). 2. **Rewrite the Equation:** Substitute \( y = x^{\frac{1}{4}} \) into the original equation: \[ y^2 + 3y - 10 = 0 \] 3. **Solve the Quadratic Equation:** \[ y^2 + 3y - 10 = 0 \] Use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 3 \), and \( c = -10 \): \[ y = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2} \] So, \[ y = \frac{4}{2} = 2 \quad \text{or} \quad y = \frac{-10}{2} = -5 \] 4. **Discard Negative Solution:** Since \( y = x^{\frac{1}{4}} \) represents a real number (assuming \( x \) is real and non-negative), we discard \( y = -5 \). 5. **Back-Substitute to Find \( x \):** \[ y = 2 \implies x^{\frac{1}{4}} = 2 \implies x = 2^4 = 16 \] 6. **Solution:** \[ x = 16 \] --- ### **Final Answers:** 1. **For \( x - 3x^{\frac{1}{2}} + 2 = 0 \):** \[ x = 1 \quad \text{and} \quad x = 4 \] 2. **For \( x^{\frac{1}{2}} + 3x^{\frac{1}{4}} - 10 = 0 \):** \[ x = 16 \]