Solving Radical Equations

By the end of this lesson, you will be able to isolate a radical and square both sides, solve what remains, and check every answer, because squaring can invent solutions that were never there.

Isolate, Then Square

$$\sqrt{x} = 5 \quad \Rightarrow \quad x = 25$$

Squaring undoes a square root, so squaring both sides removes the radical. The only requirement is that the radical is alone on its side first.

$$\sqrt{x} + 3 = 7$$

Squaring now would give $(\sqrt{x} + 3)^2 = x + 6\sqrt{x} + 9$, which still has a radical in it and is worse than what you started with. Isolate first:

$$\sqrt{x} = 4 \quad \Rightarrow \quad x = 16$$

Squaring a sum does not square the terms, which is the same $(a+b)^2$ fact from the Polynomials unit, and it is why the isolation step is not optional.

$$\textbf{isolate} \rightarrow \textbf{square} \rightarrow \textbf{solve} \rightarrow \textbf{CHECK}$$

That fourth step is not tidying. It is part of the method, for a reason that the next section is entirely about.

Why Checking Is Mandatory

Squaring is not a reversible move, and that is the whole problem.

$$x = -3 \quad \text{is false for } x = 3$$

$$x^2 = 9 \quad \text{is true for } x = 3$$

Squaring turned a false statement into a true one. So when you square both sides of an equation, the new equation can have solutions the original never had. They are called extraneous solutions, and they are not mistakes in your working: they are produced by a correct step.

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

There is no real answer, because $\sqrt{\ }$ means the positive root. But square both sides and you get $x = 4$, which looks like one. Check it: $\sqrt{4} = 2 \neq -2$. Rejected.

Only the ORIGINAL equation can settle it. Checking against the squared version proves nothing, because the squared version is exactly what invented the false answer.

A radical equal to a negative is always extraneous, and noticing that before doing any algebra saves the whole problem.

Example 1: The Whole Method

Here is every step on one equation, worked all the way through. $$\sqrt{2x + 3} = 5$$

• The radical is already alone. Square both sides. What does the left become? 2x + 3
• What does the right become? 25
• Solve 2x + 3 = 25 for x 11
• Check it. What is 2 times 11, plus 3? 25
• The square root of 25 is what? 5

Worked Example 1: Isolating First

$$\sqrt{x - 1} + 4 = 9$$

Worked Example 2: An Extraneous Solution

$$\sqrt{x + 6} = x$$

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