Example2.2.4

Is \(8\) a solution to \(x^2-5x=\sqrt{2x}+20\text{?}\)

To find out, substitute in \(8\) for \(x\) and see what happens.

\begin{align*} x^2-5x\amp=\sqrt{2x}+20\\ \substitute{8}^2-5(\substitute{8})\amp\stackrel{?}{=}\sqrt{2(\substitute{8})}+20\\ \highlight{64}-5(8)\amp\stackrel{?}{=}\sqrt{\highlight{16}}+20\\ 64-\highlight{40}\amp\stackrel{?}{=}\highlight{4}+20\\ \highlight{24}\amp\stackrel{\checkmark}{=}\highlight{24} \end{align*}

So yes, \(8\) is a solution to \(x^2-5x=\sqrt{2x}+20\text{.}\)

in-context