2.2 Solve for \( x \), without using a calculator: \( \begin{array}{ll}2.2 .1 & \sqrt[4]{32}=128 \\ 2.2 .2 & x^{\frac{1}{2}}-3 x^{\frac{1}{4}}-10=0\end{array} \)
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To solve \( \sqrt[4]{32} = 128 \), we first recognize that \( 32 \) can be expressed as \( 2^5 \). Therefore, \( \sqrt[4]{32} = (2^5)^{\frac{1}{4}} = 2^{\frac{5}{4}} \), which is not equal to \( 128 \) since \( 128 = 2^7 \). This suggests that the equation \( \sqrt[4]{32} = 128 \) is actually false. Now, let’s solve \( x^{\frac{1}{2}} - 3x^{\frac{1}{4}} - 10 = 0 \). We can make the substitution \( y = x^{\frac{1}{4}} \). Then \( x^{\frac{1}{2}} = y^2 \), so the equation transforms into \( y^2 - 3y - 10 = 0 \). Factoring this gives us \( (y - 5)(y + 2) = 0 \), thus \( y = 5 \) or \( y = -2 \). Since \( y = x^{\frac{1}{4}} \) must be non-negative, we only take \( y = 5 \), which gives us \( x^{\frac{1}{4}} = 5 \) or \( x = 625 \). So the solution is \( x = 625 \).
