When the unknown hides under a radical sign, standard algebraic techniques fall short. The radical must be eliminated — typically by raising both sides to a power. But this process can create solutions that do not satisfy the original equation.
Solving radical equations requires both algebraic manipulation and verification. Every candidate solution must be checked.
What is a Radical Equation
A radical equation contains a variable under a radical sign.
x=5
x+3=x−1
32x−1=3
The variable may appear only under the radical, or both under and outside. It may appear under multiple radicals. Each configuration requires slightly different handling, but the core strategy remains: isolate and eliminate.
Radical equations differ from expressions involving radicals with known values. Here the radicand contains the unknown, making the radical an obstacle to finding x.
Basic Strategy
The fundamental approach has four steps.
Isolate the radical on one side of the equation.
Raise both sides to the power that matches the index.
Solve the resulting equation.
Check all solutions in the original equation.
x=7
The radical is already isolated. Square both sides:
x=49
Check: 49=7. Valid.
x−2+4=9
Isolate first: x−2=5
Square: x−2=25
Solve: x=27
Check: 27−2+4=25+4=5+4=9. Valid.
Why Isolation Comes First
Squaring before isolating creates unnecessary complications.
Consider x+3=7 handled incorrectly:
(x+3)2=49
x+6x+9=49
The radical remains. The equation is now harder, not easier.
Correct approach — isolate first:
x=4
x=16
Check: 16+3=4+3=7. Valid.
Isolation ensures that raising to a power actually eliminates the radical. The techniques from operations show what happens when binomials containing radicals are squared — the radical often survives unless isolated first.
Extraneous Solutions
Squaring both sides can introduce false solutions — values that satisfy the squared equation but not the original.
x=−3
Squaring gives x=9. But 9=3=−3.
The original equation has no solution. The principal square root is never negative — a consequence of the properties of radicals. Squaring erased the negative sign and created a spurious answer.
Another example:
x+5=x−1
Square: x+5=x2−2x+1
Rearrange: x2−3x−4=0
Factor: (x−4)(x+1)=0
Candidates: x=4 and x=−1
Check x=4: 9=3 and 4−1=3. Valid.
Check x=−1: 4=2 and −1−1=−2. Invalid — 2=−2.
Only x=4 solves the original equation. The value x=−1 is extraneous.
Why Extraneous Solutions Appear
Squaring is not a reversible operation. Both 5 and −5 square to 25. When an equation is squared, information about sign is lost.
If the original equation requires a radical to equal a negative value, squaring hides this impossibility. The squared equation accepts values that the original rejects.
Similarly, if the original equation has domain restrictions — the radicand must be non-negative for even roots — squaring may produce solutions outside this domain.
Extraneous solutions are not errors in algebra. They are artifacts of an irreversible operation. The check step is not optional; it is part of the solution process.
Odd-index equations rarely produce extraneous solutions because odd roots accept all real numbers and preserve sign. But checking remains good practice.
Equations with Two Radicals
When two radicals appear, isolate one and square. Then isolate the remaining radical and square again.
x+5−x=1
Isolate one radical:
x+5=x+1
Square both sides:
x+5=x+2x+1
Simplify:
4=2x
x=2
x=4
Check: 9−4=3−2=1. Valid.
The first squaring did not eliminate all radicals — a radical remained from expanding (x+1)2 using techniques from operations. A second isolation and squaring completed the process.
Equations with Radicals on Both Sides
When each side has a single radical, square directly.
2x+3=x+7
Square:
2x+3=x+7
x=4
Check: 11=11. Valid.
This configuration rarely produces extraneous solutions because both sides have the same structure — both are principal roots, both non-negative.
More complex cases:
3x+1=x−1+2
Isolate before squaring:
3x+1−2=x−1
Square:
3x+1−43x+1+4=x−1
2x+6=43x+1
x+3=23x+1
Square again and solve. Check all candidates.
Cube Roots and Higher
Odd-index equations follow the same process but raise to the appropriate power.
3x−2=4
Cube both sides:
x−2=64
x=66
Check: 364=4. Valid.
32x+1=3x−3
Cube:
2x+1=x−3
x=−4
Check: 3−7=3−7. Valid.
Odd-index radicals handle negative radicands naturally, as explained in properties. Extraneous solutions are rare because cubing preserves sign and no principal root convention restricts outputs.
Fourth roots, sixth roots, and other even-index equations behave like square root equations — domain restrictions apply and extraneous solutions can arise.
Radical Inequalities
Inequalities involving radicals require attention to domain.
x>3
Square: x>9
But also: the radicand must satisfy x≥0 for the square root to exist.
Intersection: x>9. The domain restriction x≥0 is automatically satisfied.
x−2≤4
Square: x−2≤16, so x≤18
Domain: x−2≥0, so x≥2
Intersection: 2≤x≤18.
For inequalities, direction may reverse when both sides are squared — but only if squaring a negative. Since principal roots are non-negative, squaring inequalities involving isolated radicals typically preserves direction.
The properties page details domain restrictions. Full treatment of inequalities extends beyond radical equations into broader inequality techniques.
Verification Process
Every solution must be substituted into the original equation.
Substitute the candidate value for the variable.
Simplify both sides independently.
Compare: if they match, the solution is valid. If not, it is extraneous.
This step cannot be skipped. It is part of solving radical equations, not merely a check on arithmetic.
2x+1=x−1
After solving, suppose candidates are x=4 and x=0.
Check x=4: 9=3 and 4−1=3. Match. Valid.
Check x=0: 1=1 and 0−1=−1. No match. Extraneous.
Solution: x=4 only.
Verification protects against extraneous solutions introduced by squaring and against domain violations where radicands become negative.