Answer
$\{16\}$
Work Step by Step
We have to solve the quadratic equation:
$$x^{1/2}+3x^{1/4}-10=0.\tag1$$
We substitute $y=x^{1/4}$:
$$y^2+3y-10=0.\tag2$$
We solve equation $(2)$:
$$\begin{align*}
(y-2)(y+5)&=0\\
y-2=0&\text{ or }y+5=0\\
y=2&\text{ or }y=-5.
\end{align*}$$
We determine $x$ using the values of $y$ that we found.
$$\begin{align*}
x&=y^4\\
y&=2\Rightarrow x=2^4=16\\
y&=-5\Rightarrow x=(-5)^4=625\\
\end{align*}$$
Check the solutions:
$$\begin{align*}
x&=16\\
16^{1/2}+3(16^{1/4})-10&\stackrel{?}{=}0\\
4+3(2)-10&\stackrel{?}{=}0\\
0&=0\checkmark\\\\
x&=625\\
625^{1/2}+3(625^{1/4})-10&\stackrel{?}{=}0\\
25+3(5)-10&\stackrel{?}{=}0\\
25+15-10&\stackrel{?}{=}0\\
30&\not=0.
\end{align*}$$
We got that $x=625$ is not a solution. The solution set of the equation is:
$$\left\{16\right\}.$$
