College Algebra, Chapter 1, 1.5, Section 1.5, Problem 2

Solve the equation $\sqrt{2x} + x = 0$ by doing the following steps.

a.) Isolating the radical.


$
\begin{equation}
\begin{aligned}

\sqrt{2x} + x =& 0
&& \text{Given}
\\
\\
\sqrt{2x} =& -x
&& \text{Isolate the radical}
\\
\\
\sqrt{2} \cdot \sqrt{x} =& -x
&& \text{Divide both sides by } \sqrt{x}
\\
\\
\sqrt{2} =& \frac{-x }{\sqrt{x}}
&& \text{Apply the properties of exponent}
\\
\\
\sqrt{2} =& -x^{\left( 1 - \frac{1}{2} \right)}
&&
\\
\\
\sqrt{2} =& -x^{\frac{1}{2}}
&& \text{Square both sides}
\\
\\
(\sqrt{2})^2 =& (-x^{\frac{1}{2}})^2
&& \text{Simplify}
\\
\\
x =& 2

\end{aligned}
\end{equation}
$


b.) Squaring both sides


$
\begin{equation}
\begin{aligned}

\sqrt{2x} + x =& 0
&& \text{Given}
\\
\\
(x) =& (- \sqrt{2x})^2
&& \text{Subtract } \sqrt{2x}
\\
\\
x^2 =& 2x
&& \text{Square both sides}
\\
\\
x^2 - 2x =& 0
&& \text{Subtract } 2x
\\
\\
x( x - 2) =& 0
&& \text{Factor out } x
\\
\\
x =& 0 \text{ and } x - 2 = 0
&& \text{Zero Product Property}
\\
\\
x =& 0 \text{ and } x = 2
&& \text{Solve for } x

\end{aligned}
\end{equation}
$


c.) The solutions of the resulting quadratic equation are ______.

The solutions of the resulting quadratic equation are $x = 0$ and $x = 2$.

d.) The solution(s) that satisfy the original equation are ______.

The solution(s) that satisfy the original equation are $x = 0$ and $x = 2$.