Respect the principal square root and use the original equation to remove extraneous candidates.
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Try
A square has area 25 square centimetres. Its side length is √25 = 5 cm. The equation s² = 25 also has algebraic solution s = −5, but √25 itself denotes the nonnegative root. Distinguish these statements before solving radical equations.
Predict
Would √(x + 4) = −2 have a real solution? Squaring seems to give x = 0, but the original would say 2 = −2. Predict why the candidate fails.
Experiment
Compare √(x + 2) = x with its squared equation x + 2 = x².
Candidate x
√(x + 2)
x
Original equality?
−1
1
−1
No
2
2
2
Yes
Both candidates satisfy the squared equation, because (x − 2)(x + 1) = 0. Only 2 satisfies the original. The table shows the lost sign information.
Understand
Over the reals, √r requires r ≥ 0 and returns a nonnegative value. If √f(x) = g(x), both f(x) ≥ 0 and g(x) ≥ 0 are necessary. Squaring gives f(x) = g(x)², but without the sign condition it may admit additional candidates.
For √(2x + 3) = x, require x ≥ 0. Squaring gives x² − 2x − 3 = (x − 3)(x + 1) = 0. Candidate −1 violates the sign requirement. At x = 3, the original reads √9 = 3, so 3 is accepted.
For √(x + 5) + 1 = 4, isolate the radical first: √(x + 5) = 3. Then x + 5 = 9 and x = 4 checks. Squaring the original left side without its middle term would be another error: (a + b)² includes 2ab.
An irrational candidate should remain exact when exact roots are requested. Squaring and factoring may lead to a quadratic formula rather than rational numbers. The practice uses a genuine choice question for those roots; a close decimal is not being checked as symbolic equivalence.
Apply
Before calculating, record the radical domain and required sign of the other side. After calculating, substitute every remaining candidate into the original. In the written diagnosis, distinguish a necessary condition from a sufficient checked solution.
Continue
Polynomial factoring provides a reusable way to find candidates once denominators or radicals have been handled. Keep restrictions attached to those candidates rather than treating the transformed polynomial as the entire original problem.
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Practice and reasoning
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Solve √(x + 5) = 4 over the real numbers.
Integers, fractions and finite decimals are accepted. Separate values with commas; type none for no solutions. Expressions such as sqrt(2) are not parsed by this checker.
Solve √(x + 6) = x.
Integers, fractions and finite decimals are accepted. Separate values with commas; type none for no solutions. Expressions such as sqrt(2) are not parsed by this checker.
Solve √(2x + 1) = −3.
Integers, fractions and finite decimals are accepted. Separate values with commas; type none for no solutions. Expressions such as sqrt(2) are not parsed by this checker.
Which is the complete real solution set of √(x + 2) = x − 1?
A learner squares √(x + 1) = x − 1, finds x = 0 and 3, and keeps both. Repair the solution and explain the logical issue.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
