Linear and quadratic inequalities

An inequality states that one quantity is larger or smaller than another. Solving an inequality means finding every value that makes the statement true, not merely finding boundary values.

For example,

[ x^2-5x+6<0 ]

has boundary values (x=2) and (x=3), but its solution is the whole interval

[ 2<x<3. ]

The key question is always: where does the expression have the required sign?

Prerequisites

You should be able to:

rearrange and solve linear equations
factorise and solve quadratics
sketch graphs of functions
manipulate algebraic fractions

Inequality symbols and solution notation

The four basic symbols are:

Symbol	Meaning
(a<b)	(a) is less than (b)
(a>b)	(a) is greater than (b)
(a\leq b)	(a) is less than or equal to (b)
(a\geq b)	(a) is greater than or equal to (b)

A strict inequality, such as (x<4), excludes the boundary. An inclusive inequality, such as (x\leq4), includes it.

Solutions can be written in several equivalent ways:

[ -2<x\leq5, ]

[ x\in(-2,5], ]

[ {x\in\mathbb R:-2<x\leq5}. ]

In interval notation, a round bracket excludes an endpoint and a square bracket includes it. Infinity always takes a round bracket because (+\infty) and (-\infty) are not real endpoints:

[ x\geq3 \quad\Longleftrightarrow\quad x\in[3,\infty). ]

Solving linear inequalities

Linear inequalities are manipulated like linear equations, with one crucial exception:

Multiplying or dividing both sides by a negative number reverses the inequality sign.

This happens because multiplication by a negative reflects the number line. Since (2<5), multiplying both sides by (-1) gives (-2>-5).

Example 1: a straightforward linear inequality

Solve

[ 5x-7\leq18. ]

Add (7) to both sides:

[ 5x\leq25. ]

Divide by (5):

[ x\leq5. ]

The divisor is positive, so the sign does not change.

Example 2: dividing by a negative

Solve

[ 7-3x>19. ]

Subtract (7):

[ -3x>12. ]

Divide by (-3), reversing the inequality:

[ x<-4. ]

A quick check confirms the direction. Choose (x=-5):

[ 7-3(-5)=22>19. ]

Choose (x=0), which is outside the proposed solution:

[ 7-3(0)=7\not>19. ]

Example 3: variables on both sides

Solve

[ 4(2x-1)\geq 3x+11. ]

Expand:

[ 8x-4\geq3x+11. ]

Collect the (x)-terms:

[ 5x\geq15. ]

Therefore

[ x\geq3. ]

Compound inequalities: and versus or

The word and means the value must satisfy both conditions. The solution is the overlap, or intersection.

The word or means the value may satisfy either condition. The solution is the union.

Example 4: a double inequality

Solve

[ -5<2x+1\leq9. ]

Perform the same operation on all three parts. Subtract (1):

[ -6<2x\leq8. ]

Divide by (2):

[ -3<x\leq4. ]

This is an and statement:

[ x>-3\quad\text{and}\quad x\leq4. ]

Example 5: an or solution

Suppose factor analysis gives

[ x\leq-1\quad\text{or}\quad x>6. ]

This cannot be compressed into one chained inequality because the two intervals are separate. In set notation,

[ x\in(-\infty,-1]\cup(6,\infty). ]

Quadratic inequalities

To solve a quadratic inequality:

Rearrange so that one side is (0).
Find the roots of the corresponding quadratic equation.
Determine the sign of the quadratic in each interval.
Select the interval or intervals with the required sign.
Include roots only when the inequality contains equality.

Finding the roots gives boundaries, not the final answer.

Example 6: a factorised quadratic

Solve

[ x^2-5x+6<0. ]

Factorise:

[ (x-2)(x-3)<0. ]

The critical values are (2) and (3). They divide the number line into three intervals:

[ x<2,\qquad 2<x<3,\qquad x>3. ]

Test one value in each interval:

Interval	Test value	Sign of ((x-2)(x-3))
(x<2)	(x=0)	((-)(-)=+)
(2<x<3)	(x=2.5)	((+)(-)=-)
(x>3)	(x=4)	((+)(+)=+)

The expression is negative only between the roots, so

[ \boxed{2<x<3}. ]

Graphically, (y=x^2-5x+6) is an upward-opening parabola. It lies below the (x)-axis between its roots.

Example 7: an inclusive quadratic inequality

Solve

[ 2x^2+x-6\geq0. ]

Factorise:

[ (2x-3)(x+2)\geq0. ]

The roots are

[ x=-2\quad\text{and}\quad x=\frac32. ]

The coefficient of (x^2) is positive, so the parabola opens upwards. It is non-negative outside the roots and non-positive between them. Equality is allowed, so include both roots:

[ \boxed{x\leq-2\quad\text{or}\quad x\geq\frac32}. ]

Example 8: roots found with the quadratic formula

Solve

[ x^2+2x-7\leq0. ]

The quadratic does not factorise conveniently. Solve (x^2+2x-7=0):

[ x=\frac{-2\pm\sqrt{2^2-4(1)(-7)}}{2} =\frac{-2\pm\sqrt{32}}2 =-1\pm2\sqrt2. ]

The parabola opens upwards, so it is at or below the (x)-axis between its roots:

[ \boxed{-1-2\sqrt2\leq x\leq-1+2\sqrt2}. ]

Keep exact roots unless a decimal approximation is explicitly required.

Sign diagrams

A sign diagram records whether each factor is positive or negative in each interval. It is especially useful when an expression has several factors or a denominator.

Consider

[ (x+1)(x-2)(x-5)>0. ]

The critical values are (-1), (2) and (5). Each simple linear factor changes sign at its root.

Interval	(x+1)	(x-2)	(x-5)	Product
(x<-1)	(-)	(-)	(-)	(-)
(-1<x<2)	(+)	(-)	(-)	(+)
(2<x<5)	(+)	(+)	(-)	(-)
(x>5)	(+)	(+)	(+)	(+)

Therefore

[ \boxed{-1<x<2\quad\text{or}\quad x>5}. ]

The roots are excluded because the original inequality is strict.

Repeated factors

A factor with even multiplicity does not change sign at its root. For example,

[ (x-1)^2(x+3)<0. ]

The squared factor is non-negative on both sides of (x=1). The sign changes at (x=-3), but not at (x=1). Since ((x-1)^2>0) except at (x=1), the product has the same sign as (x+3), apart from the additional zero at (x=1). Hence

[ \boxed{x<-3}. ]

Inequalities involving brackets

Brackets may be expanded, but preserving a factorised form often makes the sign structure clearer.

Example 9: compare two products

Solve

[ (x-1)(x+4)>2(x-1). ]

Do not divide immediately by (x-1). Its sign is unknown, and it may be zero. Instead, move everything to one side:

[ (x-1)(x+4)-2(x-1)>0. ]

Factor out (x-1):

[ (x-1)\bigl((x+4)-2\bigr)>0, ]

[ (x-1)(x+2)>0. ]

The roots are (-2) and (1). The product is positive outside the roots:

[ \boxed{x<-2\quad\text{or}\quad x>1}. ]

Dividing by (x-1) at the first step could reverse the inequality when (x<1), and would discard the need to consider (x=1). Rearranging and factorising is safer.

Fractional inequalities

For a rational expression, both numerator zeros and denominator zeros are critical values.

The denominator can never equal zero, even when the inequality is inclusive.

Example 10: a linear fraction

Solve

[ \frac{x-3}{x+1}\leq0. ]

Critical values:

(x=3), where the numerator is zero
(x=-1), where the expression is undefined

Use a sign diagram:

Interval	(x-3)	(x+1)	Quotient
(x<-1)	(-)	(-)	(+)
(-1<x<3)	(-)	(+)	(-)
(x>3)	(+)	(+)	(+)

The quotient is negative between (-1) and (3), and equals zero at (x=3). It is undefined at (x=-1). Therefore

[ \boxed{-1<x\leq3}. ]

Why cross multiplication can fail

From

[ \frac{x-3}{x+1}\leq0, ]

you cannot simply multiply by (x+1) without knowing its sign. If (x+1<0), the inequality reverses. A sign diagram handles both cases at once.

Example 11: fractions on both sides

Solve

[ \frac{2}{x-1}>1. ]

Move everything to one side:

[ \frac{2}{x-1}-1>0. ]

Use a common denominator:

[ \frac{2-(x-1)}{x-1}>0, ]

[ \frac{3-x}{x-1}>0. ]

The critical values are (x=1) and (x=3). Testing intervals gives:

for (x<1), numerator positive and denominator negative, so the quotient is negative
for (1<x<3), numerator and denominator are positive, so the quotient is positive
for (x>3), numerator negative and denominator positive, so the quotient is negative

Thus

[ \boxed{1<x<3}. ]

Neither endpoint is included: (x=1) is undefined, and (x=3) makes the quotient zero while the inequality is strict.

Graphical interpretation

An inequality such as

[ f(x)>g(x) ]

asks where the graph of (y=f(x)) lies above the graph of (y=g(x)).

Example 12: comparing a line and a parabola

Solve

[ x^2-2x<3x+4. ]

The boundary points occur where the graphs meet:

[ x^2-2x=3x+4, ]

[ x^2-5x-4=0. ]

Therefore

[ x=\frac{5\pm\sqrt{41}}2. ]

Rearranging the inequality gives

[ x^2-5x-4<0. ]

This upward-opening quadratic is negative between its roots, so

[ \boxed{\frac{5-\sqrt{41}}2<x<\frac{5+\sqrt{41}}2}. ]

On this interval, the parabola (y=x^2-2x) lies below the line (y=3x+4).

Graphs provide interpretation and a valuable check, but a sketch alone may not give exact boundary values.

Two-variable inequalities and regions

An inequality in (x) and (y) usually describes a region of the coordinate plane.

To draw a region:

Replace the inequality sign with equality to find the boundary.
Draw a solid boundary for (\leq) or (\geq).
Draw a dashed boundary for (<) or (>).
Test a point not on the boundary, often ((0,0)).
Shade the side that satisfies the inequality.

Example 13: a linear half-plane

Draw the region

[ y\leq2x+3. ]

The boundary is the line (y=2x+3). It is solid because equality is included.

Test ((0,0)):

[ 0\leq2(0)+3, ]

which is true. Therefore shade the side containing the origin. Equivalently, shade on or below the line.

Example 14: simultaneous linear inequalities

Describe the region satisfying

[ x\geq0,\qquad y\geq0,\qquad x+y\leq6,\qquad y\geq x-2. ]

The first two conditions restrict the region to the first quadrant. The line (x+y=6) is included, and the region lies on or below it. The line (y=x-2) is included, and the region lies on or above it.

The feasible region is the intersection of all four half-planes. A point belongs to the region only if it satisfies every inequality.

Useful boundary intersections include:

[ y=0,\quad y=x-2 \quad\Longrightarrow\quad x=2, ]

so one vertex is ((2,0)), and

[ x+y=6,\quad y=x-2 ]

gives

[ x+(x-2)=6, ]

[ x=4,\qquad y=2. ]

Other vertices are ((0,0)) and ((0,6)). The feasible region has vertices

[ (0,0),\quad(2,0),\quad(4,2),\quad(0,6). ]

Example 15: a curved boundary

Sketch the region

[ y>x^2-4. ]

The boundary is the parabola (y=x^2-4). Draw it dashed because equality is excluded. The region consists of points above the curve.

To verify the side, test ((0,0)):

[ 0>0^2-4, ]

which is true, so the origin lies in the required region.

For

[ x^2+y^2\leq9, ]

the boundary is the circle with centre ((0,0)) and radius (3). The required region is the interior together with the circumference.

Common misconceptions

Reversing the sign after adding a negative number

The sign reverses only when both sides are multiplied or divided by a negative number. Adding or subtracting any number does not reverse it.

Reporting only the roots

Roots divide the number line into intervals. The answer to a quadratic or fractional inequality is usually one or more intervals, not just the roots.

Assuming an upward parabola is always positive

An upward-opening parabola is positive outside two distinct roots and negative between them. If it has no real roots and a positive leading coefficient, then it is positive for every real (x).

Including a denominator zero

For

[ \frac{x+2}{x-4}\geq0, ]

(x=4) is never included because the expression is undefined there. The inclusive sign does not change this.

Cancelling a factor without recording restrictions

[ \frac{(x-2)(x+1)}{x-2}, ]

cancelling gives (x+1), but the original expression still requires (x\neq2). Cancellation simplifies a formula, not its domain.

Shading by visual guesswork

Descriptions such as “above” and “below” are easy for (y)-form inequalities, but less obvious for (2x+3y>6) or circles. Substitute a test point.

Confusing and with or

For (x>1) and (x<5), take the overlap: (1<x<5).

For (x<1) or (x>5), take both separate intervals.

Checking solutions

Use these checks before accepting an answer:

Boundary check: solve the related equation and confirm every critical value has been considered.
Interval check: substitute one value from each proposed interval into the original inequality.
Endpoint check: decide separately whether each root is included, excluded, or undefined.
Graph check: compare the sign with the position of the graph relative to the (x)-axis.
Logic check: use and for an intersection and or for separate intervals.

Practice

1. Linear inequality

Solve

[ 3-5x\leq18. ]

Answer

[ -5x\leq15. ]

Dividing by (-5) reverses the sign:

[ \boxed{x\geq-3}. ]

2. Quadratic inequality

Solve

[ x^2+x-12>0. ]

Answer

[ (x+4)(x-3)>0. ]

The upward-opening quadratic is positive outside its roots:

[ \boxed{x<-4\quad\text{or}\quad x>3}. ]

3. Fractional inequality

Solve

[ \frac{x+2}{x-4}\geq0. ]

Answer

The critical values are (-2) and (4). The quotient is positive for (x<-2), negative for (-2<x<4), and positive for (x>4). Include (-2), where the quotient equals zero, but exclude (4), where it is undefined:

[ \boxed{x\leq-2\quad\text{or}\quad x>4}. ]

4. Region

State whether ((2,1)) belongs to the region

[ x\geq0,\qquad y>0,\qquad 2x+y\leq5. ]

Answer

((2,1)) satisfies (2\geq0) and (1>0). Also,

[ 2(2)+1=5\leq5. ]

Therefore ((2,1)) belongs to the region. It lies on the included boundary (2x+y=5).

Summary

To solve an inequality:

rearrange linear inequalities carefully, reversing the sign only when multiplying or dividing by a negative
rearrange polynomial and fractional inequalities so that one side is zero
find every critical value
use a graph, sign diagram, or test values to determine valid intervals
include equality roots when allowed, but never include values that make a denominator zero
write intersections using and, and separate intervals using or
for two-variable inequalities, draw the correct boundary and test a point to choose the region

What to learn next

Apply inequality restrictions when finding domains and ranges of functions.
Use inequalities with transformed curves in graph transformations.
Learn how inequalities constrain optimisation and stationary points.
Practise interpreting inequality answers in context in checking answers.