Lesson 106: Solving Radical Equations

New Concept

An equation containing a variable in a radicand is called a radical equation.

What’s next

Next, you’ll apply inverse operations, like squaring both sides, to find the value of the variable hidden inside a radical.

Property

An equation containing a variable trapped inside a radicand, like $\sqrt{x} = 9$.

Explanation

To solve these puzzles, you must free the variable! The main move is to square both sides of the equation. This is the inverse operation that undoes the square root, allowing you to find the value of the variable.

Examples

Solve $\sqrt{x} = 8$. Square both sides: $(\sqrt{x})^2 = 8^2$, so $x=64$.
Solve $\sqrt{x+3} = 5$. Square both sides: $x+3 = 25$, so $x=22$.

Property

The radical must be isolated on one side of the equation before squaring.

Explanation

To solve, get the radical by itself! Use inverse operations to move other numbers away. Once it is alone, you can square both sides of the equation. This removes the root and makes solving simple.

Examples

Solve $\sqrt{x} + 5 = 9$. Subtract 5: $\sqrt{x}=4$. Then square to get $x=16$.
Solve $4\sqrt{x} = 20$. Divide by 4: $\sqrt{x}=5$. Then square to get $x=25$.

Before you can undo the square root, you must first get it to stand alone. This example illustrates a key idea from the lesson: isolating the radical.

Example Problem

Solve the equation $\sqrt{x} - 3 = 9$.

Step-by-Step

1. To solve for $x$, we first need to isolate the radical term, $\sqrt{x}$.
2. Use the Addition Property of Equality. Add $3$ to both sides of the equation.

3. Simplify to isolate the radical.

4. Now, use the inverse operation of a square root, which is squaring. Square both sides of the equation.

5. Simplify to find the solution.

6. Finally, check the answer by substituting it back into the original equation.

Property

An answer from a derived equation that does not work in the original equation. Always check your work!

Explanation

Watch out for fakes! Squaring can create imposter answers that don't actually work. You must plug your final answer back into the original equation to check if it's a real solution or a tricky extraneous one.

Examples

Solve $\sqrt{x-2}=x-4$. Solutions are $x=3$ and $x=6$. Only $x=6$ works.
Solve $\sqrt{x+7}=-3$. Squaring gives $x=2$, but $\sqrt{9} \neq -3$. No solution.