Answer
$\{1\}$
Work Step by Step
We have to solve the quadratic equation:
$$x+7\sqrt x-8=0.\tag1$$
We substitute $y=\sqrt x$:
$$y^2+7y-8=0.\tag2$$
We solve equation $(2)$:
$$\begin{align*}
(y-1)(y+8)&=0\\
y-1=0&\text{ or }y+8=0\\
y=1&\text{ or }y=-8.
\end{align*}$$
We determine $x$ using the values of $y$ that we found:
$$\begin{align*}
x&=y^2\\
y&=1\Rightarrow x=1^2=1\\
y&=-8\Rightarrow x=(-8)^2=64.
\end{align*}$$
Check the solutions:
$$\begin{align*}
x&=1\\
1+7\sqrt 1-8&\stackrel{?}{=}0\\
1+7-8&\stackrel{?}{=}0\\
0&=0\checkmark\\\\
x&=64\\
64+7\sqrt {64}-8&\stackrel{?}{=}0\\
64+7(8)-8&\stackrel{?}{=}0\\
112&\not=0.
\end{align*}$$
We got that $x=64$ is not a solution. The solution set of the equation is:
$$\{1\}.$$
