## Precalculus (6th Edition) Blitzer

The solution of the equation is $x=625$.
Let us consider the provided equation: ${{x}^{\frac{1}{2}}}-2{{x}^{\frac{1}{4}}}-15=0$. Assume ${{x}^{\frac{1}{4}}}=t$ and put it into the equation: \begin{align} & {{x}^{\frac{1}{2}}}-2{{x}^{\frac{1}{4}}}-15=0 \\ & {{\left( {{x}^{\frac{1}{4}}} \right)}^{2}}-2{{x}^{\frac{1}{4}}}-15=0 \\ & {{t}^{2}}-2t-15=0 \\ & \left( t-5 \right)\left( t+3 \right)=0 \end{align} Further solve the equation as given below: \begin{align} & t=5,-3 \\ & {{x}^{\frac{1}{4}}}=5,-3 \end{align} When $t=5$, the value of $x$ is \begin{align} & {{x}^{\frac{1}{4}}}=5 \\ & x=625 \end{align} When $t=-3$, the value of $x$ is \begin{align} & {{x}^{\frac{1}{4}}}=-3 \\ & x=81 \end{align} But $x=81$, does not satisfy the expression ${{x}^{\frac{1}{2}}}-2{{x}^{\frac{1}{4}}}-15=0$. Thus, the solution of the equation is $x=625$.