Quadratic inequalities · Teacher reference · answers included

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Teacher reference: contains answers, conditions and proofs.

03 Quadratic inequalities · g10-03-01 · 5–10 min · TEACHER · ANSWERS

Does the middle always solve it?

What changes when the leading coefficient changes sign?

Predict: For a(x−r)(x−s)≤0, keep r,s fixed and compare positive, negative and zero a.

Compare	Computed observation
Positive coefficient Leading coefficient a = 1; Root r = -2; Root s = 3	Solution: [-2, 3]
Negative coefficient Leading coefficient a = -1; Root r = -2; Root s = 3	Solution: (−∞, -2] ∪ [3, ∞)
Zero coefficient Leading coefficient a = 0; Root r = -2; Root s = 3	Solution: (−∞, ∞)

Conclusion: Two distinct factors change sign at their roots; a negative multiplier reverses the product sign. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: Roots are real. a=0 is a constant-zero expression and is classified separately.

Justification: Order the roots u≤v. On each interval the signs of x−u and x−v are fixed. Multiplication gives the sign chart, while root substitution decides endpoints.

Ask: Have one pupil explain the graph and another explain factor signs.

Watch for: “Take the middle” assumes positive leading coefficient and two distinct roots.

Extension: Explain the repeated-root case without appealing to the graph.

Exit: Solve −(x−1)(x−4)≤0.

x≤1 or x≥4, including both roots.

Before: Factor signs and number lines · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

03 Quadratic inequalities · g10-03-02 · 5–10 min · TEACHER · ANSWERS

How many real zeros?

Can a quadratic be positive without crossing the axis?

Predict: Change c in ax²+bx+c; predict two, one or no real zeros before revealing Δ.

Compare	Computed observation
Two zeros Coefficient a = 1; Coefficient b = 0; Constant c = -1	Quadratic: two distinct real roots
A repeated root Coefficient a = 1; Coefficient b = -2; Constant c = 1	Quadratic: one repeated real root
No real zero Coefficient a = 1; Coefficient b = 0; Constant c = 1	Quadratic: no real zero
Linear degeneration Coefficient a = 0; Coefficient b = 2; Constant c = -2	Linear: one zero

Conclusion: The discriminant counts real zeros for a≠0. If a=0, the expression is linear or constant, so use its own rule. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: a,b,c are real numbers; quadratic classification requires a≠0.

Justification: For a≠0, ax²+bx+c=a[(x+b/(2a))²−Δ/(4a²)]. A square equals Δ/(4a²) at zeros, yielding two, one or zero real roots according to Δ.

Ask: Ask for a numerical invariant deciding the root count beyond the plotted samples.

Watch for: A graph outside the current viewing window is not evidence that a root is absent.

Extension: Use completing the square to explain why a repeated root does not reverse sign.

Exit: How many real zeros does x²+1 have? What sign does it have?

No real zero; it is at least 1 and therefore strictly positive.

Before: Quadratic equations and squares · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

03 Quadratic inequalities · g10-03-03 · 5–10 min · TEACHER · ANSWERS

Do the endpoints belong?

What is the difference between < and ≤ at a root?

Predict: Keep the same polynomial and change only the inequality symbol.

Compare	Computed observation
Strict inequality Coefficient a = 1; Root r = 1; Root s = 4; Inequality = Less than zero	Solution: (1, 4)
Include equality Coefficient a = 1; Root r = 1; Root s = 4; Inequality = At most zero	Solution: [1, 4]
Repeated root Coefficient a = 1; Root r = 2; Root s = 2; Inequality = Less than zero	Solution: ∅

Conclusion: A root gives exactly zero, so it belongs precisely when equality is allowed. Repeated roots need their own sign analysis. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: Use a(x−r)(x−s) with real parameters, including a=0.

Justification: At any nonroot the strict and non-strict versions have the same sign test. At each root the value is 0; comparing 0 with 0 gives its inclusion exactly.

Ask: Ask pupils to substitute each endpoint instead of memorising bracket patterns.

Watch for: Open endpoints are not approximate exclusions; they exclude exactly the root.

Extension: Classify all four inequalities for (x−2)².

Exit: Solve (x−2)²<0 and (x−2)²≤0.

The first has no solution; the second has the singleton {2}.

Before: Root signs and endpoint notation · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

03 Quadratic inequalities · g10-03-04 · 5–10 min · TEACHER · ANSWERS

A profitable value may be impossible

Which non-loss solutions are feasible in the context?

Predict: Predict the feasible set for P(x)=a(x−r)(x−s)≥0 under 0≤x≤d, then require integer x.

Compare	Computed observation
Continuous scale Coefficient a = -1; Root r = 2; Root s = 8; Capacity d = 10; Allowed sizes = Real values	Solution: [2, 8]
Capacity binds Coefficient a = -1; Root r = 2; Root s = 8; Capacity d = 6; Allowed sizes = Integer values	Solution: {2, 3, 4, 5, 6}
Repeated root is the only choice Coefficient a = -1; Root r = 3; Root s = 3; Capacity d = 6; Allowed sizes = Integer values	Solution: {3}

Conclusion: The final choice is the intersection of the sign solution, the contextual interval and, when required, the integers. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: The profit expression is a mathematical scenario; d≥1 and the integer restriction is explicit.

Justification: An x is acceptable iff it satisfies every constraint. Solving the polynomial first and intersecting cannot add an input failing any original condition.

Ask: Ask students to circle which restriction removes each rejected candidate.

Watch for: A real root outside the contextual domain is not a feasible activity size.

Extension: Find parameters with real feasible values but no feasible integer.

Exit: P(x)=−(x−2)(x−8), 0≤x≤6. Find feasible integer non-loss sizes.

2,3,4,5,6. The contextual upper bound removes 7 and 8.

Before: Polynomial signs and intersections · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.