Cracking Quadratics
Completing the square on ax² + bx + c = 0 isolates x and produces x = (−b ± √(b² − 4ac)) / 2a, the universal solver for quadratics.
A complete interactive classroom, not just a preview.
Start when you are ready to enter this Stage's 4 scenes and explore, respond, and learn as you go.
How does completing the square produce the quadratic formula, and how do you apply it?
A ball thrown upward lands where it started, but its equation x² − 5x + 6 = 0 has two answers — which one picks the right moment?
Quadratics give two solutions, but real situations often demand only one. How does the formula decide which root matters?
Step-by-step derivation of the quadratic formula by completing the square, then a worked example showing both roots and which one fits context.
Any quadratic ax² + bx + c = 0 can be solved by completing the square once, yielding x = (−b ± √(b² − 4ac)) / 2a — and context selects the valid root.
- factoring by inspection
- graphical/vertex analysis
- complex roots
- discriminant classification
- 01Two Answers, One QuestionslideSlot 1Hook
Present x² − 5x + 6 = 0 and ask whether a single number can satisfy it, given that a parabola crosses the x-axis twice.
- Quadratics can have two solutions
- Many real problems need just one
- We need a reliable way to find both
PhenomenonThe equation x² − 5x + 6 = 0 has two solutions even though it looks like one equation.
QuestionIs there a single method that finds both solutions for any quadratic?
- 02Why Not Just Guess?slideSlot 2Tension
Factoring works for friendly numbers, but fails on 2x² + 3x − 7 = 0. Predict what universal tool could always finish the job.
- Factoring is fast but limited
- Most quadratics do not factor cleanly
- We need an algebraic move that always works
PredictionA student will assume trying factors works for every quadratic.
Tempting intuitionIf a quadratic has integer roots, factoring should always find them.
- 03Completing the SquareslideSlot 3Reveal
Derive the quadratic formula by completing the square on ax² + bx + c = 0, isolating x step by step, and arrive at x = (−b ± √(b² − 4ac)) / 2a.
- Divide by a to normalize the leading coefficient
- Complete the square on bx to get (x + b/2a)²
- Isolate the squared term and take the square root
EvidenceStarting from ax² + bx + c = 0, dividing by a gives x² + (b/a)x + (c/a) = 0; completing the square yields (x + b/2a)² = (b² − 4ac) / 4a².
ConclusionEvery quadratic ax² + bx + c = 0 is solved by x = (−b ± √(b² − 4ac)) / 2a, with the ± giving the two roots.
Mechanism- 1Step 1: Divide by a and move c/a to the right side so x² + (b/a)x = −c/a.
- 2Step 2: Add (b/2a)² to both sides to form (x + b/2a)² = (b² − 4ac)/4a².
- 3Step 3: Take ± square roots of both sides and solve for x = (−b ± √(b² − 4ac)) / 2a.
- 04Using the FormulaslideSlot 4Takeaway
Apply the formula to 2x² + 5x − 3 = 0, obtain x = ½ and x = −3, then pick x = ½ for a time-based context.
- Identify a, b, c
- Compute the discriminant b² − 4ac
- Choose the root that fits the situation
TransferFor a projectile where t = 0 is launch, keep only the positive root.
Expected inferenceThe same formula works for any quadratic, and context decides which of the two roots is meaningful.
Discussion threads for a Stage aren't available yet.
