Solving Radical Equations

Solve equations containing radicals and check solutions. Grade 11 Math 20-1 Alberta.

Lesson 5.5 of Radical Expressions and Equations in Math 20-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• The Idea of Extraneous Roots. Discover why squaring both sides of an equation can introduce fake solutions called extraneous roots, and why checking your answers is never optional when solving radical equations.
• The Solving Procedure. Master the four-step procedure for solving radical equations — isolate, square, solve, check — through four fully worked examples including cases that produce quadratic equations and extraneous roots.
• Equations with Two Radicals. Learn how to handle equations containing two separate radicals by isolating and squaring twice, with a fully worked example showing each stage of the double-squaring technique.
• Word Problem — The Sphere Formula. Apply radical equation skills to a real-world context: use the surface area formula for a sphere to find radius and surface area, and see when checking for extraneous roots is and is not necessary.

Inside the lesson — a free preview

Two Equations — Two Different Answers

Why x = √(25) and x^2 = 25 Give Different Results. Consider these two equations side by side.

Equation 1: x = √(25)

The radical symbol x is defined to give only the principal (positive) root. So √(25) = 5, and the only answer is x = 5.

Equation 2: x^2 = 25

Here we ask: what numbers, when squared, give 25? Both 5 and -5 satisfy that — so x = 5 or x = -5.

Key insight: Squaring is not a perfect inverse of taking a square root. Squaring loses sign information — and that is exactly what causes extraneous roots.

When There Is No Solution

Example — √(x - 2) + 3 = 0. Isolate the radical:

√(x - 2) = -3

The left side √(x - 2) is a square root — it is always greater than or equal to zero. The right side is -3, which is negative.

No value of x can make this equation true.

Answer: No solution.

Key Takeaway — Spot No-Solution Cases Early. Whenever you isolate a radical and find it equals a negative number, the equation has no solution — without doing any further work.

If you square both sides anyway, you will get a value of x that looks like a solution but does not actually work. That value would be an extraneous root.

Worked example: One Real Solution

Example 1: Solve √(x + 2) - 5 = 0. Step 1 — Isolate the radical. Add 5 to both sides:

√(x + 2) = 5

Step 2 — Square both sides:

(√(x + 2))^2 = 5^2

x + 2 = 25

Step 3 — Solve for x. Subtract 2 from both sides:

x = 23

Step 4 — Check x = 23 in the original equation:

√(x + 2) - 5 = 0

√(23 + 2) - 5 = √(25) - 5 = 5 - 5 = 0 ✓

Answer: x = 23

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