The learning goal
For the real equation sqrt(x+6)=x, squaring gives candidates 3 and -2, but the original requires a nonnegative right side and only 3 survives. Squaring can discard sign information. Suitable nonnegative restrictions can preserve equivalence; omitting those conditions does not justify keeping every candidate.
A principal square root is not a two-root equation
sqrt(9) is the nonnegative principal root, 3. The equation x²=9 has solutions 3 and -3. They are different mathematical objects. Writing sqrt(9)=±3 adds a sign choice that the expression did not contain.
The OpenStax section on radical equations explains that even powers can introduce values that do not solve the original equation and requires checking there. The examples and variation below are original editorial exercises, not a study-effectiveness experiment.
Name three objects separately
Domain
Values for which the original expressions are defined.
Candidates
Values solving the transformed equation that still require selection.
Original solutions
Defined values that make the original two sides equal.
Solve sqrt(x+6)=x from start to finish
First require x+6≥0, hence x≥-6. Because the square root on the left is nonnegative, equality also requires x≥0. The combined necessary restriction can therefore be retained as x≥0.
Squaring gives x+6=x². Rearrange to x²-x-6=0 and factor as (x-3)(x+2)=0, producing candidates x=3 and x=-2. Calling them candidates avoids announcing both as solutions to the original.
| Step | Statement | Reason |
|---|---|---|
| Domain | x≥-6 | The radicand must be nonnegative |
| Sign | x≥0 | The right side equals a nonnegative root |
| Square | x+6=x² | Remove the radical while retaining the sign condition |
| Candidates | x=3 or x=-2 | Solve the transformed quadratic |
| Original check | Keep x=3 only | Check the original equality |
Why does -2 pass the square but fail the original?
At 3, the original left side is sqrt(9)=3 and the right side is 3. At -2, the left side is sqrt(4)=2 but the right side is -2. Yet the squared equation gives 4=4. The value solves the transformed equation without solving the original problem.
Substituting only into the squared equation checks candidate generation. It does not recover the sign information discarded by the transformation.
| Candidate x | Original left side | Original right side | Decision |
|---|---|---|---|
| 3 | sqrt(9)=3 | 3 | Keep |
| -2 | sqrt(4)=2 | -2 | Reject |
Locate the first unjustified inference
The squared equation has two roots, so both 3 and -2 solve the original.
The square supplies two candidates; the original sign condition and substitution reject -2.
Change the right side to -x: same candidates, different answer
Now solve sqrt(x+6)=-x. The radicand still requires x≥-6, while the right-side sign condition becomes x≤0. Squaring still gives x+6=x² and the same candidates 3 and -2.
In the new original equation, -2 gives 2 on both sides, while 3 gives 3 on the left and -3 on the right. Only -2 remains. Identical squared equations do not force identical original solution sets; the difference is exactly the sign hidden by squaring.
Original task: sqrt(x + 6) = x
Variant: sqrt(x + 6) = -x
For each, write the domain, the sign condition on the right side,
the candidates after squaring, and a substitution into the original equation.
Explain why identical squared equations do not force identical solution sets.Record the point where reverse reasoning becomes invalid
Do not stop at “forgot to check roots” in an error log. Record the original domain, signs, implication direction and the condition that rejects each candidate. Explain the reasons before reading the solution again. Remembering 3 without explaining why -2 fails leaves the critical step unresolved.
As in the converse and contrapositive exercise, the direction of reasoning matters. Use the step-by-step self-explanation record to attach a reason to each line instead of recopying the whole answer.
Conditions for completing the radical-equation task
- Real-domain conditions for the original are stated.
- The principal square root has not been replaced by ±.
- Roots of the square are candidates with necessary conditions retained.
- Every kept or rejected value has an original-equation justification.
- A sign-changed variation is reconsidered rather than answered by copying the earlier result.
Common questions
If x≥0 was retained, must I still substitute 3?
A correct derivation that rigorously retains equivalence conditions can justify its selection that way. For learning and handoff review, direct substitution is still a clear check that can catch arithmetic mistakes.
Does squaring always introduce an extraneous root?
No. It can add candidates, but whether any are actually extraneous depends on the equation and its conditions. Possibility is not inevitability.
Why not say sqrt(9)=±3?
The radical denotes the principal square root, which is 3. The two values ±3 solve x²=9; a value of an expression and a solution set are different objects.
