Unlike the previous problems squaring both sides once isn’t sufficient to eliminate the square roots from the problem. So, once we get the remaining root on one side by itself, as we did in the previous step, we need to square both sides once again.

Doing that gives,

\[\begin{align*}{\left( {2\sqrt {1 - x} } \right)^2} & = {\left( {3x + 2} \right)^2}\\ 4\left( {1 - x} \right) & = 9{x^2} + 12x + 4\\ 9{x^2} + 16x & = 0\end{align*}\] Show Step 3

This is just a quadratic equation and we know how to solve it so let’s do that.

\[x\left( {9x + 16} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = 0,\,\,\,\,\,x = - \frac{{16}}{9}\]

As shown we have two solutions to the quadratic we got from the first step.

Show Step 4

We’re not done with this problem. Recall from the notes that the solution process we used here has the unfortunate side effect of sometimes introducing values that are not solutions to the original equation.

So, to finish this out we need to check both of the potential solutions from the previous step in the original equation (recall it’s important to check the potential solutions in the original equation).

\[x = 0:\hspace{0.25in}1 + \sqrt {1 - 0} \mathop = \limits^? \sqrt {2\left( 0 \right) + 4} \hspace{0.25in} \to \hspace{0.25in}1 + \sqrt 1 \mathop = \limits^? \sqrt 4 \hspace{0.25in} \to \hspace{0.25in}2 = 2\hspace{0.25in}{\mbox{OK}}\] \[x = - \frac{{16}}{9}:\hspace{0.25in}1 + \sqrt {1 - \left( { - \frac{{16}}{9}} \right)} \mathop = \limits^? \sqrt {2\left( { - \frac{{16}}{9}} \right) + 4} \hspace{0.25in} \to \hspace{0.25in}1 + \sqrt {\frac{{25}}{9}} \mathop = \limits^? \sqrt {\frac{4}{9}} \hspace{0.25in} \to \,\,\,\,\,\,\frac{8}{3} \ne \frac{2}{3}\hspace{0.25in}{\mbox{NOT}}\,\,{\mbox{OK}}\]

Only one of the potential solutions work out and so the original equation has a single solution : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0}}\) .