Radical expressions & equations
The idea
A radical expression contains a root — √x, ∛x, or higher — and it ties straight back to exponents, because √x is x^(1/2) and the exponent laws you already trust keep working. That link explains rules like √(ab) = √a × √b for nonnegative a and b, and it is why simplifying √72 into 6√2 is really just factoring out perfect squares. Radicals appear wherever the Pythagorean theorem does, so distances, diagonals, and falling-object times all speak this language.
Solving a radical equation means isolating the radical and raising both sides to the matching power. Squaring both sides, though, is not fully reversible: it makes 3 and −3 look identical, so it can smuggle in extraneous solutions — candidates that satisfy the squared equation but not the original. Checking every candidate in the original equation is part of the method, not an optional courtesy. The misconception to retire is √(a² + b²) = a + b; with a = 3 and b = 4 the left side is 5 while the right side is 7, so a root never splits across addition.
Worked example
Solve √(2x + 7) = x − 4, and identify any extraneous solutions.
- Before squaring, note that the square root on the left is never negative, so any genuine solution needs x − 4 ≥ 0, that is, x ≥ 4.
- The radical is already isolated, so square both sides: 2x + 7 = (x − 4)² = x² − 8x + 16.
- Bring everything to one side: x² − 10x + 9 = 0, which factors as (x − 1)(x − 9) = 0, giving candidates x = 1 and x = 9.
- Test x = 1 in the original equation: the left side is √9 = 3 but the right side is 1 − 4 = −3. They disagree, so x = 1 is extraneous — squaring erased the sign difference.
- Test x = 9: the left side is √(18 + 7) = √25 = 5 and the right side is 9 − 4 = 5. This one holds, and it also respects the x ≥ 4 requirement spotted at the start.
Answer. The only solution is x = 9; the candidate x = 1 is extraneous.
Check your understanding
- Why does squaring both sides of an equation sometimes create solutions the original equation rejects, and at exactly which moment is the information lost?
- How does writing √x as x^(1/2) let the exponent laws explain why √a × √b = √(ab) for nonnegative numbers?
- What changes when the equation uses a cube root instead of a square root, and why do extraneous solutions stop appearing?
- How could you tell before doing any algebra that √(2x + 7) = x − 4 cannot have a solution smaller than 4?
Build the foundations first
Radical expressions & equations builds on these concepts. If any feel shaky, start there.