Solving Radical Equations

By the end of this lesson, you will be able to solve radical equations with square and cube roots, isolate radicals before squaring/cubing, handle equations with two radicals, and always check for extraneous solutions.

The Big Idea

This is an extension of the Algebra 1 radical equations lesson. We now handle more complex cases: equations with two radicals requiring two squarings, equations where the quadratic from squaring requires the quadratic formula, and rational exponent equations solvable by raising both sides to a power.

The Core Procedure (Review)

1. Isolate the radical
2. Raise both sides to the power equal to the index ($2$ for square root, $3$ for cube root)
3. Solve the resulting equation
4. Check all solutions — extraneous roots are always possible for even-index radicals

Example 1 — Two Radicals

Solve: $\sqrt{x + 5} - \sqrt{x - 3} = 2$

Isolate one radical: $\sqrt{x+5} = \sqrt{x-3} + 2$

Square both sides:

$$x + 5 = (x - 3) + 4\sqrt{x-3} + 4$$

$$x + 5 = x + 1 + 4\sqrt{x-3}$$

$$4 = 4\sqrt{x-3} \implies 1 = \sqrt{x-3}$$

Square again: $x - 3 = 1 \implies x = 4$

Check: $\sqrt{9} - \sqrt{1} = 3 - 1 = 2$ ✓

Example 2 — Quadratic from Squaring

Solve: $\sqrt{2x + 3} = x - 1$

Square both sides: $2x + 3 = x^2 - 2x + 1$

$$0 = x^2 - 4x - 2 \implies x = \frac{4 \pm \sqrt{24}}{2} = 2 \pm \sqrt{6}$$

Check $x = 2 + \sqrt{6} \approx 4.45$:

$\sqrt{2(4.45)+3} = \sqrt{11.9} \approx 3.45$ and $4.45 - 1 = 3.45$ ✓

Check $x = 2 - \sqrt{6} \approx -0.45$:

$\sqrt{2(-0.45)+3} = \sqrt{2.1} \approx 1.45$ but $-0.45 - 1 = -1.45$

$1.45 \neq -1.45$ (square root can't equal negative) ✗ — extraneous.

Solution: $x = 2 + \sqrt{6}$ only.

Example 3 — Cube Root Equation

Solve: $\sqrt[3]{5x - 2} = 3$

Cube both sides: $5x - 2 = 27 \implies x = \frac{29}{5}$

No check needed for cube roots (cubing is reversible — no extraneous solutions for odd indices).

Example 4 — Rational Exponent Equation