Solve for \( x \) : (a) \( x^{\frac{1}{2}}-3 x^{\frac{1}{4}}+2=0 \)

To solve the equation \( x^{\frac{1}{2}} - 3 x^{\frac{1}{4}} + 2 = 0 \), we can make a substitution to simplify it. Let \( y = x^{\frac{1}{4}} \). Then, we know that \( x^{\frac{1}{2}} = y^2 \). Substituting, we get the equation: \[ y^2 - 3y + 2 = 0. \] Next, we can factor this quadratic equation: \[ (y - 1)(y - 2) = 0. \] Setting each factor to zero gives us: 1. \( y - 1 = 0 \) which means \( y = 1 \) 2. \( y - 2 = 0 \) which means \( y = 2 \) Now we substitute back for \( x \): 1. If \( y = 1 \), then \( x^{\frac{1}{4}} = 1 \) which implies \( x = 1^4 = 1 \). 2. If \( y = 2 \), then \( x^{\frac{1}{4}} = 2 \) which implies \( x = 2^4 = 16 \). Finally, the solutions for \( x \) are: \( x = 1 \) and \( x = 16 \).