Algebra - Equations with Radicals

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Section 2.10 : Equations with Radicals

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3. Solve the following equation.

\[7 = \sqrt {39 + 3x} - x\]

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The first step here is to square both sides. However, before we do that we need to get the root on one side by itself.

\[7 + x = \sqrt {39 + 3x} \]

Now we can square both sides to get,

\[\begin{align*}{\left( {7 + x} \right)^2} & = {\left( {\sqrt {39 + 3x} } \right)^2}\\ {x^2} + 14x + 49 & = 39 + 3x\\ {x^2} + 11x + 10 & = 0\end{align*}\] Show Step 2

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

\[\left( {x + 10} \right)\left( {x + 1} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - 10,\,\,\,\,\,x = - 1\]

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

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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).

\[x = - 10:\hspace{0.25in}7\mathop = \limits^? \sqrt {39 + 3\left( { - 10} \right)} - \left( { - 10} \right)\hspace{0.25in} \to \hspace{0.25in}7\mathop = \limits^? \sqrt 9 + 10\hspace{0.25in} \to \hspace{0.25in}7 \ne 13\hspace{0.25in}{\mbox{NOT}}{\mbox{ OK}}\] \[x = - 1:\hspace{0.25in}7\mathop = \limits^? \sqrt {39 + 3\left( { - 1} \right)} - \left( { - 1} \right)\hspace{0.25in} \to \hspace{0.25in}7\mathop = \limits^? \sqrt {36} + 1\hspace{0.25in} \to \hspace{0.25in}7 = 7\hspace{0.25in}{\mbox{OK}}\]

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