When solving equations that involve radicals, begin by asking: is there an $x$ under the square root? The answer to this question will determine the way the problem is approached. If there is not an $x$ under the square root—if only numbers are under the radicals—the problem can be solved much the same way as if it had no radicals. 

Solving Radical Equations without Variables Under the Radical Symbol

Example

Solve this equation:

$\sqrt 2 x+5=7-\sqrt 3 x$

First, isolate the variable on one side of the equation:

$\begin{aligned} \sqrt 2 x+\sqrt 3 x+5-5&=7-5-\sqrt 3 x + \sqrt 3 x \\ \sqrt 2x+\sqrt 3x&=7-5 \end{aligned}$

Next, since both terms on the left hand side of the equation contain $x$, factor out the $x$. (Remember that $\sqrt 2 $ and $\sqrt 3$ are unlike terms and cannot be combined.)

$x(\sqrt 2+\sqrt 3)=2$

Now divide both sides of the equation by $(\sqrt 2+\sqrt 3)$ to solve for $x$:

$x=\dfrac 2 {\sqrt 2 + \sqrt 3}$

The key thing to note about problems like this is that both sides of the equation do not have to be squared. $2\sqrt{2}$ may look complicated, but it is just a number—it functions in the equation just the way that the number $10$, or $\frac {1}{3}$, or $\pi$ would.

Steps to Solve a Radical Equation with a Variable Under the Radical

If there is an $x$, or variable, under the square root, the problem must be approached differently. In this case, both sides must be squared to get rid of the radical. However, squaring both sides can introduce extraneous solutions (i.e., false answers), so it is important to check the answers after solving. If no answer checks out, then the solution is "no solution."

Example

Solve this equation:

$\sqrt{6x-2}-3=7$

First, isolate the radical:

$\begin{aligned} \sqrt{6x-2}-3+3&=7+3 \\ \sqrt{6x-2}&=10 \end{aligned}$

Now, to undo the radical symbol (square root), square both sides of the equation (recall that squaring a square root removes the radical):

$\begin{aligned} \left(\sqrt{6x-2}\right)^2&=(10)^2 \\ 6x-2&=100 \end{aligned}$

Finally, solve the remaining equation:

$\begin{aligned} 6x-2+2&=100+2 \\ 6x&=102 \\ \dfrac{6x}{6}&=\dfrac{102}{6} \\ x&=17 \end{aligned}$

Now, let's go back and check our answer:

$\begin{aligned} \sqrt {6(17)-2}-3&=7 \\ \sqrt {6(17)-2}&=10 \\ \left(\sqrt {6(17)-2}\right)^2&=(10)^2 \\ 6(17)-2&=100 \\ 6(17)&=102 \\ 17&=17 \end{aligned} $

This is a true statement. Therefore, $x=17$ is a valid solution to the equation $\sqrt{6x-2}-3=7$.

Solving Radical Equations with False Solutions

Solve this equation:

$-5-\sqrt{10x-2}=5$

Isolate the radical symbol (square root):

$\begin{aligned} -5-\sqrt{10x-2}+5&=5+5 \\ -\sqrt{10x-2}&=10 \\ (-1) \cdot (-\sqrt{10x-2})&=(-1) \cdot (10) \\ \sqrt{10x-2}&=-10 \end{aligned} $

 Now, square both sides of the equation, and solve:

$\begin{aligned} \left( \sqrt{10x-2} \right) ^2&= (-10) ^2 \\ 10x-2&=100 \\ 10x-2+2&=100+2 \\ 10x&=102 \\ \dfrac{10x}{10}&=\dfrac{102}{10} \\ x&=10.2 \end{aligned}$

Once we check this result, however, we discover that $\sqrt {100}=-10$. This is incorrect, because the square root is defined to be only the positive root, $10$. This means that $10.2$ is an extraneous solution. Because it is the only answer we found, the answer to this problem is "no solution."

This problem demonstrates how important it is to check your solutions whenever you square both sides of an equation.