LEARN · EXPLAIN · REVISE

Quadratic signs and solution sets

Read the idea, work independently, then explain what changed.

高一必修第一册(A版).pdf · 2.3 · PDF 57 / printed page 50

Revisit first: Roots and completeness of solutions

TOPIC 01

Quadratic signs and solution sets

Use roots to partition a line, then retain endpoint and domain conditions.

What you will be able to explain

Justify the sign on each interval instead of memorising an unqualified inside/outside rule.
Intersect an algebraic solution set with modelling and domain restrictions.

Roots partition sign intervals

Solving an equation gives boundary candidates, not the inequality solution set.

A continuous quadratic can change sign only at its roots; factor signs justify each interval.

a(x-r_1)(x-r_2)

For r₁<r₂, list the two factor signs in each region.
$(-)(-),\quad(+)(-),\quad(+)(+)$
Multiply by the sign of a.
$a>0:+,-,+;\quad a<0:-,+,-$

When it applies

Distinct real roots are assumed for this three-region chart; repeated and absent roots need separate handling.

Counterexample: −(x−1)(x−3) is positive inside the roots, not outside.

Coordinate a sign chart, a parabola and interval notation.

Endpoints and excluded values

An open or closed endpoint encodes a mathematical condition, not a drawing preference.

A weak inequality may include a root, but it never restores a forbidden denominator.

\frac{x-1}{x+2}\ge0,\quad x\ne-2

Separate numerator zeros and denominator exclusions.
$x=1;\quad x\ne-2$
Test all intervals, then include only allowed equality points.
$(-\infty,-2)\cup[1,\infty)$

When it applies

Do not multiply by an expression of unknown sign without splitting cases.

Counterexample: Multiplying by x can reverse an inequality when x<0.

Show roots as closed or open dots only after checking the original domain.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: How do the positive intervals change when a changes sign or the two roots merge?

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

Target f(x) ≥ 0. Solution: (−∞,-1] ∪ [2,∞). Repeated roots and a=0 are treated separately.

Numerical data for the diagram; printable
Probe input	Value	Satisfies target?
-2	4	✓
-1	0	✓
0.5	-2.25	×
2	0	✓
3	4	✓

Explain: Read signs from factors and compare strict versus weak endpoints.

Transfer: Construct a weak inequality whose solution is exactly one point.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Foundation#Worked example

Solve the strict inequality.

(x-1)(x-4)>0

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Partition at the two roots.

Hint 2

A product is positive when both factors have the same sign.

Worked solution

Locate the boundaries and signs.
$1,4;\quad+,-,+$
Exclude roots because the inequality is strict.
$x<1\text{ or }x>4$

(−∞,1)∪(4,∞).

Checks and common pitfalls: The endpoints make the product zero, not positive.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Solve and justify inclusion of endpoints.

-x^2+4x-3\ge0

First step provided — complete the reasoning:Use the positive-leading form.

(x-1)(x-3)\le0

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Factor out the negative sign.

Hint 2

Multiply the inequality by −1 and reverse it.

Worked solution

Use the positive-leading form.
$(x-1)(x-3)\le0$
Choose the inside region with equality.
$1\le x\le3$

[1,3].

Checks and common pitfalls: The leading sign reverses the usual sign pattern.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

Find every k making the expression strictly positive for every real x.

x^2-2x+k>0

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use the global minimum, not a few test inputs.

Hint 2

Complete the square around x=1.

Worked solution

Expose the smallest value.
$x^2-2x+k=(x-1)^2+k-1$
Strict positivity includes the vertex input.
$k-1>0\Rightarrow k>1$

k>1.

Checks and common pitfalls: k=1 fails at x=1, even though all other inputs give positive values.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

How many integers satisfy the inequality?

x^2<9

Count integer x only.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Solve over the reals first.

Hint 2

The interval is open at −3 and 3.

Worked solution

Convert the square bound.
$-3<x<3$
List the permitted integers.
$-2,-1,0,1,2$

Checks and common pitfalls: Neither endpoint is allowed.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Explain exactly what changes if > is replaced by ≥.

(x-2)(x-5)>0

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The interval signs remain unchanged.

Hint 2

Only equality at roots is added.

Worked solution

The strict solution uses two open rays.
$(-\infty,2)\cup(5,\infty)$
The weak inequality adds precisely the two roots.
$(-\infty,2]\cup[5,\infty)$

Add x=2 and x=5, and nothing else.

Checks and common pitfalls: Changing endpoint inclusion does not reverse the interval signs.

Reasoning checklist · self / teacher assessment

The strict solution uses two open rays.
The weak inequality adds precisely the two roots.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Solve the inequality and explain why an interval is not obtained.

(x-2)^2\le0

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

A real square cannot be negative.

Hint 2

Only equality is possible.

Worked solution

Combine nonnegativity with the given bound.
$0\le(x-2)^2\le0$
The square must vanish.
$x=2$

The singleton {2}.

Checks and common pitfalls: A double root does not create a negative region.

Reasoning checklist · self / teacher assessment

Combine nonnegativity with the given bound.
The square must vanish.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Solve and justify the answer for every real input.

x^2+x+1>0

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Complete the square.

Hint 2

The remaining constant is positive.

Worked solution

Complete exactly.
$x^2+x+1=(x+1/2)^2+3/4$
Every output is at least 3/4.
$x^2+x+1\ge3/4>0$

All real x.

Checks and common pitfalls: No roots alone is insufficient without the sign of the function.

Reasoning checklist · self / teacher assessment

Complete exactly.
Every output is at least 3/4.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Solve both restrictions simultaneously.

x^2-5x+6\le0,\quad x>2

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Solve the quadratic restriction first.

Hint 2

Intersect [2,3] with x>2.

Worked solution

Factor and select the weak inside interval.
$(x-2)(x-3)\le0\Rightarrow x\in[2,3]$
Apply the additional strict restriction.
$x\in(2,3]$

(2,3].

Checks and common pitfalls: A valid endpoint for one condition may fail the other.

Reasoning checklist · self / teacher assessment

Factor and select the weak inside interval.
Apply the additional strict restriction.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Solve without multiplying by an unknown-sign denominator.

\frac{x-1}{x+2}\ge0

x≠−2.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Mark −2 as forbidden and 1 as a zero.

Hint 2

Test the three sign intervals.

Worked solution

The quotient is positive outside the two critical values.
$x<-2\text{ or }x>1$
Include only the allowed zero.
$x\in(-\infty,-2)\cup[1,\infty)$

(−∞,−2)∪[1,∞).

Checks and common pitfalls: The weak inequality does not admit the denominator zero.

Reasoning checklist · self / teacher assessment

The quotient is positive outside the two critical values.
Include only the allowed zero.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Find the real domain of this function.

y=\sqrt{-x^2+6x-5}

Use the real square root.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The radicand must be nonnegative.

Hint 2

Factor the radicand with its negative sign.

Worked solution

Convert the domain requirement.
$-(x-1)(x-5)\ge0$
Use the closed inside interval.
$x\in[1,5]$

[1,5].

Checks and common pitfalls: A zero radicand is allowed; a negative one is not.

Reasoning checklist · self / teacher assessment

Convert the domain requirement.
Use the closed inside interval.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

A student divides by x and concludes x<2. Repair the argument.

x(x-2)<0

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The sign of x is not known.

Hint 2

A negative product needs opposite signs.

Worked solution

Partition at both roots.
$0,2;\quad+,-,+$
Select the negative region and reject zero endpoints.
$0<x<2$

0<x<2.

Checks and common pitfalls: x=−1 refutes the student answer: the product is positive.

Reasoning checklist · self / teacher assessment

Partition at both roots.
Select the negative region and reject zero endpoints.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

A rectangle has perimeter 20 m. For which positive side lengths is its area at least 21 m²? State all possible shorter-side lengths too.

2(x+y)=20,\quad xy\ge21

Positive real side lengths, all four sides fenced.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Eliminate y, then distinguish an arbitrary side from the shorter side.

Hint 2

The area condition becomes (x−3)(x−7)≤0.

Worked solution

Combine the model with positive lengths.
$y=10-x,\quad x\in[3,7]$
The shorter side is min(x,10−x).
$3\le\min(x,y)\le5$

Sides x and 10−x with 3≤x≤7; shorter side lies in [3,5].

Checks and common pitfalls: A second interpretation is required after solving the algebraic inequality.

Reasoning checklist · self / teacher assessment

Combine the model with positive lengths.
The shorter side is min(x,10−x).

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Find every a such that the expression is nonnegative for all real x; account for degenerate cases.

(a-1)x^2+2x+1\ge0

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

A negative leading coefficient fails for large |x|; a=1 is linear.

Hint 2

For a>1 use the vertex or Δ≤0.

Worked solution

Exclude nonpositive leading cases.
$a<1: f\to-\infty;\quad a=1:2x+1$
Positive leading coefficient requires no crossing below zero.
$4-4(a-1)\le0\Rightarrow a\ge2$

a≥2.

Checks and common pitfalls: At a=2 the square (x+1)² attains zero, which is permitted.

Reasoning checklist · self / teacher assessment

Exclude nonpositive leading cases.
Positive leading coefficient requires no crossing below zero.

Think first. Reveal a hint when the class is ready.

End-of-lesson check and correction

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Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Diagnose regions	Use the negative-leading and repeated-root diagnostics.
8	Build signs	Students multiply factor signs on three intervals.
10	Endpoints and domains	Pairs contrast strict, weak, radical and rational restrictions.
12	Model and quantify	Groups solve rectangle feasibility and then the all-real parameter condition.
5	Exit interval	Require a sign reason and correct endpoint notation.

Question sequence

What information is missing if we only solve the equation?
Does “for every x” permit checking only three values?

Board development

Multiply signs before drawing the solution.
$a(x-r_1)(x-r_2)$
Intersect with original restrictions.
$S_{\mathrm{valid}}=S_{\mathrm{sign}}\cap D$

Expected correct and incorrect responses

“The square is nonnegative, so strict negativity is impossible” is a complete reason.
“Include −2 because ≥ allows equality” overlooks the denominator.

Follow-up questions

Which endpoint would survive after adding a denominator?

Remediation

For inside/outside errors, rebuild a two-factor sign chart instead of recalling a slogan.

Exit check

Solve (x−1)²≥0 and (x−1)²>0; explain the single-point difference.

Assessment checklist and sample responses

2 marks: sign regions and equality.
Full: (2,3] after intersection; incomplete: [2,3].
2 marks: model interpretation.
Full: arbitrary side [3,7], shorter side [3,5]; partial: no shorter-side restriction.