Extraneous solution after squaring: test the original

An extraneous solution after squaring is a candidate that satisfies the transformed equation but fails the original one. In the paper problem here, squaring produces minus two and three as candidates. Only three survives substitution into the original equation. The repair is to keep the original condition visible, rather than accept every value produced by a later algebraic step.

Use this exact equation over the real numbers: √(x + 6) = x. The square-root sign denotes the principal square root: the nonnegative number whose square equals the quantity underneath it. Everything below concerns this explicitly defined problem. It is not a report that either calculator linked later malfunctioned or produced an incorrect answer.

Record what the original equation requires

The quantity under the radical is x + 6. For the left side to be defined as a real square root, that quantity must be nonnegative. Thus x must be at least minus six. Keep this domain condition beside the original equation before changing its form.

Equality imposes another condition. The left side is nonnegative by the principal-root convention, so the right side x must also be nonnegative. This is stronger than merely requiring x to be at least minus six. A number can leave the radical defined and still be unable to equal its nonnegative value.

For example, minus two leaves x + 6 equal to four, which has a defined principal square root. That fact does not make minus two a solution. The original equation requires the value of that root to equal x, not merely to exist. Keeping existence and equality separate is the first useful check in this problem.

Square both sides while retaining the earlier condition

If a real value satisfies the original equality, squaring its two equal sides gives x + 6 = x². This supplies a necessary equation for every original solution. Move all terms to one side to obtain x² − x − 6 = 0. Neither operation erases the earlier requirement that an original solution has x at least zero.

The danger is in reversing the implication without that condition. Equal real numbers have equal squares, but equal squares do not always come from equal numbers. Two and minus two both square to four while remaining different numbers. Squaring can therefore make two sides agree that did not agree before.

Write a note saying candidate equation beside the squared form if that helps preserve its role. A value found from it still needs to satisfy the original radical equation. The later polynomial is useful progress, without becoming permission to replace the original question by a different one.

Factor the complete candidate equation

The polynomial factors as x² − x − 6 = (x − 3)(x + 2). Multiplying the factors gives x² + 2x − 3x − 6, which simplifies to the same expression. This direct expansion checks the signs in the factorization before any candidate is accepted.

A product of two real factors is zero only when at least one factor is zero. Here that gives x − 3 = 0 or x + 2 = 0. The two algebraic candidates are therefore three and minus two. We have exhausted the roots of this factored candidate equation, rather than searched a few integers and guessed that no others exist.

Both candidates genuinely satisfy the squared equation. For three, x² is nine and x + 6 is nine. For minus two, x² is four and x + 6 is four. Finding that agreement does not expose a factoring mistake. It shows precisely why checking only the transformed equation would leave the original problem unfinished.

Substitute minus two into the unchanged original

Return to √(x + 6) = x, with the radical and its principal-root meaning intact. Substituting minus two makes the left side √4, which equals positive two. The right side remains minus two. Two does not equal minus two, so this candidate fails the original equation.

The failed check is exact. It does not depend on a rounding tolerance, a numerical display or how many digits were retained. Nor is the radical undefined here: its argument is four. The failure is that its nonnegative value differs from the negative value required on the other side.

This is the extraneous candidate introduced when the sign distinction was lost in squaring. You could have excluded it earlier using the necessary condition x at least zero. Showing the original substitution nevertheless makes the rejection inspectable and explains why the squared equation had allowed it.

Substitute three before reporting the answer

Putting three into the original equation gives √9 on the left and three on the right. The principal square root of nine is three, so the equality holds. Three also meets the earlier domain and sign conditions. It is an original solution, rather than merely a value that survived a sign filter.

That final distinction matters. A nonnegative candidate need not solve every equation simply because it passes a necessary sign condition. In this problem, direct substitution supplies the additional equality check. Retain that check beside the accepted value instead of writing only that the negative candidate was discarded.

Completeness now follows from the full chain. Every original solution must satisfy the squared polynomial. That polynomial has only the two candidates already found. One fails the original equality and the other passes. Therefore three is the only real solution to the stated original equation.

Distinguish a root symbol from a two-solution equation

The expression √4 denotes two under the principal-root convention. It does not denote the pair two and minus two. By contrast, the equation y² = 4 has both values as solutions. These are different requests, and writing an extra plus-or-minus sign into √4 would change the meaning used in our original problem.

You can see the same issue by beginning with x = 3 and squaring to get x² = 9. Minus three satisfies the second equation but not the first. The new candidate did not reveal that the original equality had secretly contained another answer. It appeared because the transformed condition retained less sign information.

For a negative x, taking the principal square root of x² returns its nonnegative magnitude, not that negative x. At x equal to minus two, √(x²) is two. This concrete control explains why cancelling a square and a square-root sign without checking signs can create an invalid reversal.

Keep an undefined expression separate from a false equality

As another control, put minus seven into the original radical's argument. Then x + 6 is minus one, which has no real principal square root in the convention of this problem. That is a domain failure. It differs from the minus-two case, whose radical is defined but whose equality fails.

Do not combine those explanations into a vague statement that every rejected value gives an error. A useful diagnostic says what failed: the expression was outside the stated real domain, or the two defined sides were unequal. Keeping those categories distinct helps another reader reproduce the particular check instead of guessing what a display might have meant.

Check the equation that was actually asked

A worksheet can report zero after subtracting the sides of the squared equation at minus two: four minus four is zero. Subtracting the sides of the original equation instead gives two minus minus two, or four. The two checks concern different expressions. A successful residual check for one cannot silently certify the other.

If an answer sheet retains both candidates, preserve its factorization and add the missing original substitutions. There is no need to pretend every earlier step was wrong. If it writes √4 as minus two, correct that principal-root value and then compare the sides again. Identify the changed step so the repair remains traceable.

A different original equation requires a different check. Replacing the right side by minus x would change the sign requirement and the equality being tested. Keep that as a new problem with its own record, rather than borrowing the final answer three from the equation solved here.

Keep other information checks in their own scope

The Fibonacci parity-pattern guide retains two consecutive parity labels because one label does not determine the next. Its result concerns a specified sum recurrence and indexed terms. This problem instead tracks which real candidates still satisfy an original equality after a transformation loses sign information.

The last digit power-sum guide keeps only the decimal information its final-digit question needs. Here the missing sign remains necessary to the original question, so it cannot be discarded at acceptance. No power cycle or last-digit lookup supplies the radical equation's solution.

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The Texas Instruments TI-30XS blue five-pack offer describes a 16-digit LCD, four-line display and battery/solar power. Its sole gallery shows five blue-and-white TI-30XS MultiView calculators. Those are offer disclosures and an illustration, without an independently counted delivery, runtime test or observed execution of either candidate check.

Neither selected account establishes this equation as supplied content or documents a performed calculation of either candidate.

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