Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as

$\begin{array}{l}\sqrt{3x+18}\hfill&=x\hfill \\ \sqrt{x+3}\hfill&=x - 3\hfill \\ \sqrt{x+5}-\sqrt{x - 3}\hfill&=2\hfill \end{array}$

Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions.

### A General Note: Radical Equations

An equation containing terms with a variable in the radicand is called a radical equation.

### How To: Given a radical equation, solve it.

1. Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.
2. If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an nth root radical, raise both sides to the nth power. Doing so eliminates the radical symbol.
3. Solve the remaining equation.
4. If a radical term still remains, repeat steps 1–2.
5. Confirm solutions by substituting them into the original equation.

### Example 6: Solving an Equation with One Radical

Solve $\sqrt{15 - 2x}=x$.

### Solution

The radical is already isolated on the left side of the equal side, so proceed to square both sides.

$\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ {\left(\sqrt{15 - 2x}\right)}^{2}\hfill&={\left(x\right)}^{2}\hfill \\ 15 - 2x\hfill&={x}^{2}\hfill \end{array}$

We see that the remaining equation is a quadratic. Set it equal to zero and solve.

$\begin{array}{l}0\hfill&={x}^{2}+2x - 15\hfill \\ \hfill&=\left(x+5\right)\left(x - 3\right)\hfill \\ x\hfill&=-5\hfill \\ x\hfill&=3\hfill \end{array}$

The proposed solutions are $x=-5$ and $x=3$. Let us check each solution back in the original equation. First, check $x=-5$.

$\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ \sqrt{15 - 2\left(-5\right)}\hfill&=-5\hfill \\ \sqrt{25}\hfill&=-5\hfill \\ 5\hfill&\ne -5\hfill \end{array}$

This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.

Check $x=3$.

$\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ \sqrt{15 - 2\left(3\right)}\hfill&=3\hfill \\ \sqrt{9}\hfill&=3\hfill \\ 3\hfill&=3\hfill \end{array}$

The solution is $x=3$.

### Try It 5

Solve the radical equation: $\sqrt{x+3}=3x - 1$

Solution

Solve $\sqrt{2x+3}+\sqrt{x - 2}=4$.

### Solution

As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.

$\begin{array}{ll}\sqrt{2x+3}+\sqrt{x - 2}\hfill& =4\hfill & \hfill \\ \sqrt{2x+3}\hfill& =4-\sqrt{x - 2}\hfill & \text{Subtract }\sqrt{x - 2}\text{ from both sides}.\hfill \\ {\left(\sqrt{2x+3}\right)}^{2}\hfill& ={\left(4-\sqrt{x - 2}\right)}^{2}\hfill & \text{Square both sides}.\hfill \end{array}$

Use the perfect square formula to expand the right side: ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$.

$\begin{array}{ll}2x+3\hfill& ={\left(4\right)}^{2}-2\left(4\right)\sqrt{x - 2}+{\left(\sqrt{x - 2}\right)}^{2}\hfill & \hfill \\ 2x+3\hfill& =16 - 8\sqrt{x - 2}+\left(x - 2\right)\hfill & \hfill \\ 2x+3\hfill& =14+x - 8\sqrt{x - 2}\hfill & \text{Combine like terms}.\hfill \\ x - 11\hfill& =-8\sqrt{x - 2}\hfill & \text{Isolate the second radical}.\hfill \\ {\left(x - 11\right)}^{2}\hfill& ={\left(-8\sqrt{x - 2}\right)}^{2}\hfill & \text{Square both sides}.\hfill \\ {x}^{2}-22x+121\hfill& =64\left(x - 2\right)\hfill & \hfill \end{array}$

Now that both radicals have been eliminated, set the quadratic equal to zero and solve.

$\begin{array}{ll}{x}^{2}-22x+121=64x - 128\hfill & \hfill \\ {x}^{2}-86x+249=0\hfill & \hfill \\ \left(x - 3\right)\left(x - 83\right)=0\hfill & \text{Factor and solve}.\hfill \\ x=3\hfill & \hfill \\ x=83\hfill & \hfill \end{array}$

The proposed solutions are $x=3$ and $x=83$. Check each solution in the original equation.

$\begin{array}{l}\sqrt{2x+3}+\sqrt{x - 2}\hfill& =4\hfill \\ \sqrt{2x+3}\hfill& =4-\sqrt{x - 2}\hfill \\ \sqrt{2\left(3\right)+3}\hfill& =4-\sqrt{\left(3\right)-2}\hfill \\ \sqrt{9}\hfill& =4-\sqrt{1}\hfill \\ 3\hfill& =3\hfill \end{array}$

One solution is $x=3$.

Check $x=83$.

$\begin{array}{l}\sqrt{2x+3}+\sqrt{x - 2}\hfill&=4\hfill \\ \sqrt{2x+3}\hfill&=4-\sqrt{x - 2}\hfill \\ \sqrt{2\left(83\right)+3}\hfill&=4-\sqrt{\left(83 - 2\right)}\hfill \\ \sqrt{169}\hfill&=4-\sqrt{81}\hfill \\ 13\hfill&\ne -5\hfill \end{array}$

The only solution is $x=3$. We see that $x=83$ is an extraneous solution.

### Try It 6

Solve the equation with two radicals: $\sqrt{3x+7}+\sqrt{x+2}=1$.

Solution