Secondary 3 Additional Mathematics | Singapore G3
Equations & Inequalities

Why Solving an Equation and Solving an Inequality Are Different Problems

An equation asks where two quantities are equal. An inequality asks where one remains larger or smaller over an entire region.

That difference changes the answer.

Solve:

x²−5x+6=0.

The answer is two points:

x=2 or x=3.

But solve:

x²−5x+6<0.

The answer is an interval:

2<x<3.

The same quadratic appears in both questions. The mathematical job is different.

The current Singapore G3 Additional Mathematics syllabus includes quadratic root conditions and related line–curve intersection/tangency conditions, simultaneous equations in two variables where one equation is linear, and quadratic inequalities represented on a number line.

SEAB 2027 G3 Additional Mathematics syllabus (K341) →

The Topic Job

Translate conditions into algebra, solve for all admissible values, and distinguish point solutions from interval solutions.

Quadratic Root Conditions

For:

ax²+bx+c=0,

the discriminant is:

Δ=b²−4ac.

Condition	Roots	Graph meaning
Δ>0	two distinct real roots	two x-intercepts
Δ=0	two equal real roots	touches x-axis
Δ<0	no real roots	no x-intercepts

The discriminant is not only an equation tool. It counts intersections.

Line–Curve Conditions

If a line and a curve intersect, their y-values are equal at the intersection.

So the standard route is:

equate the two expressions → obtain a quadratic → use Δ to count intersections.

Two intersections: Δ>0. Tangent: Δ=0. No intersection: Δ<0.

Worked Example 1 — Parameter for Tangency

The line y=2x+k is tangent to y=x²−4x+7. Find k.

At intersection:

x²−4x+7=2x+k.

Rearrange:

x²−6x+(7−k)=0.

Tangent means equal roots:

36−4(7−k)=0.

So:

36−28+4k=0 → 8+4k=0 → k=−2.

The geometry word “tangent” became one algebra condition.

Simultaneous Equations: One Linear Equation Is a Doorway

Suppose:

y=2x+1
x²+y²=25.

The linear equation expresses y in terms of x. Substitute it into the nonlinear equation:

x²+(2x+1)²=25.

Now the two-variable problem becomes one quadratic equation in x.

linear relation → substitution → one-variable equation → back-substitute.

Worked Example 2 — Two Intersection Points

Solve:

y=x+1
x²+y=7.

Substitute y=x+1:

x²+x+1=7
x²+x−6=0
(x+3)(x−2)=0.

So x=−3 or 2.

Back-substitute:

• x=−3 → y=−2;
• x=2 → y=3.

Solutions:

(−3,−2) and (2,3).

Do not stop after finding x. A simultaneous system asks for ordered pairs unless the question says otherwise.

Quadratic Inequalities Are Sign Problems

Solve:

x²−5x+6<0.

Factorise:

(x−2)(x−3)<0.

The critical points 2 and 3 divide the number line into three intervals:

• x<2;
• 2<x<3;
• x>3.

The product is negative only when the two factors have opposite signs. That happens between the roots.

Therefore:

2<x<3.

Why You Cannot Solve an Inequality by “Moving the Sign”

An inequality is not an equation with a different symbol.

The sign can reverse when multiplying or dividing both sides by a negative quantity.

With a quadratic factorisation, the safer method is usually to identify roots and analyse intervals rather than performing uncontrolled division by expressions whose sign changes with x.

Graph View of a Quadratic Inequality

For y=x²−5x+6, the roots are 2 and 3 and the parabola opens upward.

The graph lies below the x-axis between the roots and above it outside the roots.

f(x)<0 means ask where the graph lies below the x-axis.

This gives a geometric check for the sign diagram.

Strict and Inclusive Inequalities

If:

(x−2)(x−3)≤0,

the roots themselves are allowed because the expression can equal zero.

Answer:

2≤x≤3.

For < or >, use open endpoints. For ≤ or ≥, include roots if they satisfy the original inequality.

When the Leading Coefficient Is Negative

Consider:

−(x−1)(x−4)>0.

The negative sign reverses the sign pattern of (x−1)(x−4).

Or multiply both sides by −1 and reverse the inequality:

(x−1)(x−4)<0.

So:

1<x<4.

Parameter Inequalities

Suppose x²+kx+9>0 for every real x.

The parabola opens upward. To remain strictly above the x-axis, it must have no real roots:

k²−36<0.

Therefore:

−6<k<6.

This connects back to the canonical Quadratic Functions owner, which handles the deeper “always positive/negative” structure.

Extraneous Solutions Can Appear

When equations involve surds, denominators or logarithms, algebraic transformations can create candidates that are not allowed in the original problem.

Examples:

• squaring can lose sign information;
• multiplying by a denominator can hide a forbidden denominator zero;
• logarithm arguments must remain positive.

candidate solution ≠ accepted solution until it survives the original conditions.

The Earliest Weak Link

What you see	Likely weak link
Uses Δ=0 for two intersections	root-count meaning not secure
Finds x-values but not y-values in simultaneous equations	forgets system asks for ordered pairs
Quadratic inequality answer gives roots only	equation-solving habit applied to interval problem
Flips inequality sign randomly	negative multiplication/division rule not understood
Includes endpoints for strict inequality	</> versus ≤/≥ distinction weak
Accepts every transformed solution	original-domain check missing

Common Mistakes to Repair

• Solving f(x)<0 by setting f(x)=0 and stopping. Roots are boundaries, not usually the complete answer.
• Dividing an inequality by an expression containing x. Its sign may be unknown.
• Forgetting to reverse inequality after multiplying/dividing by a negative number.
• Forgetting back-substitution in simultaneous equations.
• Using discriminant conditions without first forming the correct quadratic.
• Ignoring original restrictions after transformations.

Retrieval Check

1. What discriminant condition gives two equal real roots?
2. What discriminant condition corresponds to a line missing a quadratic curve?
3. Why does a linear equation help solve a simultaneous nonlinear system?
4. Solve (x−1)(x−5)<0.
5. Solve (x−1)(x−5)≥0.
6. Why do roots of a quadratic inequality usually mark boundaries rather than the complete answer?

Transfer Set

1. Find k if y=x+k is tangent to y=x²−3x+4.
2. Solve y=3x−1 and x²+y=11 simultaneously.
3. Solve x²−7x+10≤0.
4. Solve −2x²+5x+3>0.
5. Find the values of p for which x²+px+4=0 has no real roots.
6. A candidate solution arose after squaring a surd equation. Explain exactly what check must be performed before accepting it.

Answer outline — open only after attempting

1. Equate: x²−4x+(4−k)=0. Tangent: 16−4(4−k)=0 → 4k=0 → k=0.
2. x²+3x−1=11 → x²+3x−12=0. Solve for x, then y=3x−1.
3. (x−2)(x−5)≤0 → 2≤x≤5.
4. −2x²+5x+3=−(2x+1)(x−3). Inequality >0 gives −1/2<x<3.
5. p²−16<0 → −4<p<4.
6. Substitute every candidate into the original unsquared equation and reject any that fails or violates its domain.

Independent Checks

• For a simultaneous solution, substitute the ordered pair into both original equations.
• For a quadratic inequality, test one sample value from each interval.
• Sketch the parabola mentally to confirm the sign pattern.
• For a tangency condition, verify the resulting quadratic really has Δ=0.
• For transformed equations, check candidates against the original domain.

For Parents and Tutors — What This Topic Is Really Testing

Students often know the algebra but misidentify the job.

Before allowing any calculation, ask:

• Is the answer supposed to be a point, an ordered pair or an interval?
• What does the discriminant mean in this particular geometry?
• What are the boundary values?
• What sign is required between or outside those boundaries?
• Did any algebraic step introduce candidates that need checking?

The teaching target is condition → mathematical object → complete admissible solution.

Where This Connects Next

• Quadratic Functions
• Coordinate Geometry
• Surds
• Exponential & Logarithmic Functions
• Secondary 3 A-Math Complete Map
• Additional Mathematics Directory

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Bukit Timah Tutor Mathematics
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