Surd equations require two kinds of control: algebraic manipulation and logical checking. Squaring can remove a square root, but it can also create candidate solutions that do not satisfy the original equation.
1. State the real-domain condition first
For √(x+3), require x≥−3. For √(5−2x), require x≤5/2. These conditions describe where the original real-valued equation exists.
2. Isolate the radical before squaring
Solve √(x+5)=x−1. The left side is nonnegative, so any solution must also satisfy x−1≥0, hence x≥1.
Square: x+5=(x−1)²=x²−2x+1. Thus x²−3x−4=0, giving x=4 or x=−1.
The condition x≥1 removes −1. Checking x=4 in the original gives 3=3. Therefore x=4.
3. Why extraneous solutions appear
If a=b, then a²=b². The reverse is not always true: a²=b² allows a=b or a=−b. Squaring therefore loses sign information.
This is why every candidate produced after squaring should be checked in the original equation.
4. Worked example with two radicals
Solve √(x+1)+1=√(x+6).
The domain requires x≥−1. Isolate by squaring once: x+6=x+2+2√(x+1), so 4=2√(x+1). Thus √(x+1)=2 and x=3.
Check: √4+1=3 and √9=3. Therefore x=3.
5. Do not square more than necessary
If a radical can be isolated and simplified before another squaring step, do so. Repeated expansion creates more algebra and more opportunities for extraneous candidates.
6. Equations with constant surds
Solve x+√2=3√2. This needs no squaring: x=2√2. A surd does not automatically require a radical-equation method.
7. Compare exact forms before decimalising
If an equation simplifies to x=√12−√3, simplify first: √12=2√3, so x=√3. Converting both terms to rounded decimals would make a simple exact cancellation less visible.
8. A candidate can satisfy the squared equation but fail the original
For √(x+2)=−x, the left side is nonnegative, so require x≤0 as well as x≥−2. Squaring gives x+2=x², or x²−x−2=0, so x=2 or −1.
x=2 violates x≤0 and fails the original. x=−1 gives 1=1. Therefore x=−1.
9. Common mistakes
- Squaring before isolating the radical.
- Forgetting the nonnegative sign of a principal square root.
- Accepting every root of the squared equation.
- Using decimal approximations when exact simplification is available.
10. Practice
- Solve √(x+6)=x.
- Solve √(2x+3)=x.
- Solve √(x+4)=2.
- Solve √(x+2)=2−x.
- Simplify the exact solution x=√27−√12.
11. Answers
- x≥0; x+6=x² gives x=3 or −2; only x=3 works.
- x≥0; 2x+3=x² gives x=3 or −1; only x=3 works.
- x+4=4, so x=0.
- Require x≤2 and x≥−2. Squaring gives x+2=(2−x)², so x²−5x+2=0. Candidates (5±√17)/2; only (5−√17)/2 satisfies x≤2.
- 3√3−2√3=√3.
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