# Solving Radical Equations

**Radical equations**are equations that contain variables in the

**radicand**(the expression under a radical symbol), such as

[latex]\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}[/latex]

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.

- Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.
- 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
*n*th root radical, raise both sides to the*n*th power. Doing so eliminates the radical symbol. - Solve the remaining equation.
- If a radical term still remains, repeat steps 1–2.
- Confirm solutions by substituting them into the original equation.

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

Solve [latex]\sqrt{15 - 2x}=x[/latex].### Solution

The radical is already isolated on the left side of the equal side, so proceed to square both sides.[latex]\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}[/latex]

We see that the remaining equation is a quadratic. Set it equal to zero and solve.
[latex]\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}[/latex]

The proposed solutions are [latex]x=-5[/latex] and [latex]x=3[/latex]. Let us check each solution back in the original equation. First, check [latex]x=-5[/latex].
[latex]\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}[/latex]

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 [latex]x=3[/latex].
[latex]\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}[/latex]

The solution is [latex]x=3[/latex].
### Try It 5

Solve the radical equation: [latex]\sqrt{x+3}=3x - 1[/latex] Solution### Example 7: Solving a Radical Equation Containing Two Radicals

Solve [latex]\sqrt{2x+3}+\sqrt{x - 2}=4[/latex].### Solution

As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.[latex]\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}[/latex]

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

Now that both radicals have been eliminated, set the quadratic equal to zero and solve.
[latex]\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}[/latex]

The proposed solutions are [latex]x=3[/latex] and [latex]x=83[/latex]. Check each solution in the original equation.
[latex]\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}[/latex]

One solution is [latex]x=3[/latex].
Check [latex]x=83[/latex].
[latex]\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}[/latex]

The only solution is [latex]x=3[/latex]. We see that [latex]x=83[/latex] is an extraneous solution.
### Try It 6

Solve the equation with two radicals: [latex]\sqrt{3x+7}+\sqrt{x+2}=1[/latex]. Solution## Licenses & Attributions

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