g10-05-15 · Core topics
Squaring can create a false root
Why does squaring √(x+2)=x produce two candidates but only one solution?
1 MIN · PREDICT ON PAPER
Predict whether x=−1 and x=2 satisfy the original equation before checking either graph.
Discuss or write first. No typed answer is required to continue.
At x=2: √(x+2)=2, x=2; original equation holds, squared equation holds.
2 MIN · EXPLAIN AND REVISE
Which condition does your claim need?
Compare at least two cases. Write a reason, then check the explanation together.
Reveal explanation and conditions
Squaring gives x²=x+2, with candidates −1 and 2. The original square root is nonnegative, so its right side must satisfy x≥0. Substitution rejects −1 and accepts 2.
Conditions: Real radicand requires x≥−2; equality with the principal square root additionally requires x≥0.
Mathematical justification: For real A≥0, √A=B is equivalent to A=B² together with B≥0. Here x²−x−2=(x−2)(x+1)=0; retaining x≥0 leaves x=2 only.
1 MIN · EXIT CHECK
Solve √(x+2)=x and identify the extraneous candidate.
Check after answering
Solution x=2; candidate x=−1 is extraneous because 1≠−1.
Teacher guidance and extension
Before: Real-number operations and reading coordinates; write a prediction on paper.
Ask: Write an implication arrow for squaring, then ask what added condition makes it equivalence.
Watch for: Squaring an equation always preserves exactly its solutions.
Challenge: Compare the intersections of y=√(x+2), y=x, and y=x² versus y=x+2.
Suggested timing: predict 1 min, compare 3 min, explain 2 min, exit check 1 min. Observations can suggest a claim; they do not prove it for every input.
Optional local notebook
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