Paul's Online Notes
Paul's Online Notes
Home / Algebra / Solving Equations and Inequalities / Equations with Radicals
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2.10 : Equations with Radicals

4. Solve the following equation.

\[x = 1 + \sqrt {2x - 2} \]

Show All Steps Hide All Steps

Start Solution

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.

\[x - 1 = \sqrt {2x - 2} \]

Now we can square both sides to get,

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

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

\[\left( {x - 1} \right)\left( {x - 3} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = 1,\,\,\,\,\,x = 3\]

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

Hint : Recall that the solution process used here can, and often does, introduce values that are not in fact solutions to the original equation!
Show Step 3

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).

\[x = 1:\hspace{0.25in}1\mathop = \limits^? 1 + \sqrt {2\left( 1 \right) - 2} \hspace{0.25in} \to \hspace{0.25in}1\mathop = \limits^? 1 + \sqrt 0 \hspace{0.25in} \to \hspace{0.25in}1 = 1\hspace{0.25in}{\mbox{OK}}\] \[x = 3:\hspace{0.25in}3\mathop = \limits^? 1 + \sqrt {2\left( 3 \right) - 2} \hspace{0.25in} \to \hspace{0.25in}3\mathop = \limits^? 1 + \sqrt 4 \hspace{0.25in} \to \hspace{0.25in}3 = 3\hspace{0.25in}{\mbox{OK}}\]

Both of the potential solutions work out and so the original equation has a two solutions : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 1\,\,\,{\mbox{and }}x = 3}}\) .