Pure 2.5 - Linear and quadratic inequalities | Edexcel A-Level Mathematics

An inequality describes every value in a solution set, rather than just the roots of an equation. You will solve linear, quadratic and fractional inequalities, express their intervals clearly, and represent solutions on number lines and coordinate graphs.

Inequalities describe solution sets

Can multiplying an inequality by its unknown make you lose half the answer? In $3/x<2$ , every negative $x$ works, yet an incautious multiplication by $x$ suggests only $x>3/2$ . We will recover the missing interval by keeping track of signs.

You already know

An equation asks where two values are equal. An inequality compares their order: $3<5$ , but $-3>-5$ . Multiplication by a negative number reverses the order on a number line.

Start with a linear case. Addition and subtraction preserve the inequality sign. Multiplication or division by a positive number preserves it; by a negative number reverses it. These rules are not supplied in the formula booklet.

Solution set

The solution set contains all allowed inputs making the inequality true. For real inputs, $\{x\in\mathbb R:x\ge-4\}=[-4,\infty)$ . A square bracket includes a finite endpoint; a round bracket excludes it. Infinity is never an included endpoint.

You may abbreviate the set builder to $\{x:x\ge-4\}$ when the domain is clear. Membership $x\in A$ says that $x$ belongs to $A$ ; $x\notin A$ says it does not. If every member of $A$ belongs to $B$ , then $A\subseteq B$ . The empty set $\varnothing$ contains no solutions.

Worked example 1: reversing the sign

Question. Solve $-3x+7\le19$ .

Plan. Isolate the $x$ -term, then divide by its negative coefficient.

\begin{aligned} -3x&\le12 &&\text{subtract }7\\ x&\ge-4 &&\text{divide by }-3\text{ and reverse} \end{aligned}

Why this step? Dividing by a negative reflects the ordered values, reversing their order.

Answer. $x\ge-4$ , or $[-4,\infty)$ .

Check. $x=-4$ gives equality; $x=0$ gives $7\le19$ , while $x=-5$ gives the false statement $22\le19$ .

Change only $\le$ to $<$ . The boundary value stays the same, but it must be excluded.

Worked example 2: a strict endpoint

Question. Solve $-3x+7<19$ .

Plan. Perform the same rearrangement, preserving strictness.

\begin{aligned} -3x&<12 &&\text{subtract }7\\ x&>-4 &&\text{divide by }-3 \end{aligned}

Why this step? Reversing the direction does not introduce equality.

Answer. $x>-4$ , or $(-4,\infty)$ .

Check. At $x=-4$ , $19<19$ is false, so the endpoint is excluded.

A bounded interval uses and: $-1<x\le5$ means $x>-1$ and $x\le5$ , the intersection $\{x:x>-1\}\cap\{x:x\le5\}$ . Separate intervals use or, the union $\cup$ . For instance, $x<-2$ or $x>3$ is $(-\infty, -2)\cup(3, \infty)$ .

Watch out

Wrong: “ $x<-2$ and $x>3$ ” for two outside intervals. No number meets both conditions. Use “or” to collect both intervals.

Keep the comparison's direction clear when you collect variables on one side.

Use roots to split a quadratic inequality

For $(x+2)(x-3)$ , the factors change sign at $-2$ and $3$ . On each interval between these points, neither factor changes sign, so their product keeps one sign. An upward parabola is positive outside two distinct roots and negative between them.

[DIAGRAM: asset_slug: p1_2_5_solve_linear_and_quadratic_inequalities_in_a_single_variable_and_interpret_such_inequalities_graphically__diagram_01; asset_name: Quadratic signs and a number line; caption: Positive outside the roots and negative between; description: The upward parabola y equals (x plus 2)(x minus 3) crosses at minus 2 and 3. A number line uses open endpoints and highlights x less than minus 2 or x greater than 3 for the strict positive inequality.]

This sign pattern is not a formula in the booklet. Use a sketch or test a point in each interval instead of memorising “between” without checking the leading coefficient. A downward parabola reverses the signs.

Worked example 3: a strict quadratic inequality

Question. Solve $x^2-x-6>0$ .

Plan. Find the roots, then choose where the upward parabola is positive.

\begin{aligned} x^2-x-6&=(x+2)(x-3) &&\text{factorise}\\ x&=-2\quad\text{or}\quad3 &&\text{boundary roots} \end{aligned}

Why this step? Roots divide the line into intervals of constant sign; they are boundaries, not the whole answer.

Answer. $x<-2$ or $x>3$ .

Check. Test $x=-3$ , $0$ , $4$ : the expression is $6$ , $-6$ , $6$ , respectively. The roots themselves give zero and are excluded.

Change only $>$ to $\ge$ . The positive intervals remain, and equality adds both roots.

Worked example 4: including the roots

Question. Solve $x^2-x-6\ge0$ .

Plan. Use the same roots and include the endpoints.

\begin{aligned} (x+2)(x-3)&\ge0 &&\text{factorise}\\ x&\le-2\quad\text{or}\quad x\ge3 &&\text{outside, including roots} \end{aligned}

Why this step? At each root the product is exactly zero, which now satisfies the inequality.

Answer. $(-\infty, -2]\cup[3, \infty)$ .

Check. Both $x=-2$ and $x=3$ give $0\ge0$ .

Repeated roots need care: $(x-1)^2\ge0$ holds for all real $x$ , while $(x-1)^2<0$ has solution set $\varnothing$ . A repeated root need not change the sign. Likewise, $-(x-1)^2\le0$ holds everywhere.

Watch out

Wrong: “The roots are $-2$ and $3$ , so $-2<x<3$ .” That interval gives a negative upward quadratic. Choose the sign requested, using a sketch or a test value.

This time the inequality asks for non-positive values of a different upward quadratic.

Reduce brackets and fractions safely

Brackets usually need expansion and collection. Fractions need an additional decision: is the multiplier definitely positive? If not, multiplying without cases may reverse the inequality for some values and silently lose solutions.

Worked example 5: brackets first

Question. Solve $(x-1)(x+4)\ge6$ .

Plan. Expand, move everything to one side and factorise the new quadratic.

\begin{aligned} x^2+3x-4&\ge6 &&\text{expand}\\ x^2+3x-10&=(x+5)(x-2)\ge0 &&\text{subtract }6\text{ and factorise} \end{aligned}

Why this step? The roots must come from the expression after the comparison with $6$ , not the original brackets.

Answer. $x\le-5$ or $x\ge2$ .

Check. At either endpoint the original product is $6$ ; at $x=0$ , it is $-4$ , which is excluded.

Change only the comparison to strict $>$ ; the algebra is unchanged, but the endpoints leave the set.

Worked example 6: strict bracket inequality

Question. Solve $(x-1)(x+4)>6$ .

Plan. Reduce to the same quadratic and use open endpoints.

\begin{aligned} (x-1)(x+4)-6&>0 &&\text{subtract }6\\ (x+5)(x-2)&>0 &&\text{expand and factorise} \end{aligned}

Why this step? Subtracting the same constant preserves order, so the new inequality is equivalent.

Answer. $x<-5$ or $x>2$ .

Check. $x=3$ gives $14>6$ ; $x=2$ gives $6>6$ , which is false.

For the opening fraction, $x\ne0$ . Multiplication by $x$ has an unknown sign, but multiplication by $x^2>0$ is safe for every allowed input. Keep $x=0$ excluded after solving.

Worked example 7: an unknown-sign denominator

Question. Solve $3/x<2$ .

Plan. Exclude zero and multiply by the positive square $x^2$ .

\begin{aligned} 3x&<2x^2 &&\text{multiply by }x^2>0\\ x(2x-3)&>0 &&\text{rearrange and factorise} \end{aligned}

Why this step? A positive multiplier preserves the inequality for negative and positive $x$ alike.

Answer. $x<0$ or $x>3/2$ .

Check. $x=-1$ gives $-3<2$ ; $x=1$ gives the false statement $3<2$ ; $x=2$ gives $1.5<2$ .

Change only $<$ to $\le$ . The finite non-zero boundary joins the answer, but zero remains forbidden.

Worked example 8: an exclusion survives equality

Question. Solve $3/x\le2$ .

Plan. Multiply by $x^2>0$ , solve, then restore the exclusion.

\begin{aligned} x(2x-3)&\ge0 &&\text{positive-square multiplication}\\ x&\le0\quad\text{or}\quad x\ge\frac32 &&\text{quadratic signs} \end{aligned}

Why this step? The transformed polynomial is defined at zero, but the original fraction is not.

Answer. $x<0$ or $x\ge3/2$ .

Check. $x=3/2$ gives equality; $x=0$ cannot be substituted in the original.

The opening puzzle's missing negative interval has now been recovered. This safe-multiplier method is not supplied in the formula booklet.

Watch out

Wrong: $3/x<2\Rightarrow3<2x$ for all $x$ . That implication assumes $x>0$ . Use $x^2$ , or split into positive and negative cases and reverse the sign in the negative case.

Try a non-strict fractional inequality with a different numerator and comparison value.

Interpret curve below line inequalities

“Curve below line” compares two outputs at the same input. For $y=x^2-2x-3$ and $y=x+1$ , it means $x^2-2x-3<x+1$ . Subtracting the line's output turns that comparison into the sign of a single quadratic.

[DIAGRAM: asset_slug: p1_2_5_solve_linear_and_quadratic_inequalities_in_a_single_variable_and_interpret_such_inequalities_graphically__diagram_02; asset_name: Curve below a line; caption: Compare outputs between the intersection inputs; description: The curve y equals x squared minus 2x minus 3 is below the line y equals x plus 1 for x strictly between minus 1 and 4. The two intersections mark the excluded endpoints.]

This comparison is not the same as asking where the curve is negative. Both graphs may be positive while one is lower: at $x=7/2$ , their outputs are $9/4$ and $9/2$ , so the positive curve is below the positive line.

Worked example 9: strictly below

Question. Find where $y=x^2-2x-3$ lies below $y=x+1$ .

Plan. Form the output inequality, collect terms and use its roots.

\begin{aligned} x^2-2x-3&<x+1 &&\text{curve below line}\\ x^2-3x-4&=(x+1)(x-4)<0 &&\text{subtract the line} \end{aligned}

Why this step? The difference is negative exactly when the curve's output is smaller.

Answer. $-1<x<4$ .

Check. At $x=0$ , $-3<1$ ; at $x=5$ , $12<6$ is false.

Change only “below” to “on or below”. The intersection inputs are now included.

Worked example 10: on or below

Question. Find where $y=x^2-2x-3$ lies on or below $y=x+1$ .

Plan. Use a non-strict comparison with the same difference quadratic.

\begin{aligned} (x+1)(x-4)&\le0 &&\text{subtract outputs}\\ -1&\le x\le4 &&\text{between the roots} \end{aligned}

Why this step? Equality means both outputs are equal: the graphs intersect there.

Answer. $[-1, 4]$ .

Check. At $x=-1$ , both outputs are $0$ ; at $x=4$ , both are $5$ .

Graph comparisons appear later when you find intersections and areas between curves.

Watch out

Wrong: solving $x^2-2x-3<0$ for “curve below the line”. The correct comparison subtracts $x+1$ , giving $x^2-3x-4<0$ .

Apply that translation to a different pair of graphs.

Represent inequalities graphically

An inequality in $x$ and $y$ describes a region of the plane. Replace its sign with equality to find the boundary; then decide whether the boundary belongs and which side to shade. A strict inequality uses a dotted boundary, and an inclusive inequality uses a solid boundary.

[DIAGRAM: asset_slug: p1_2_5_solve_linear_and_quadratic_inequalities_in_a_single_variable_and_interpret_such_inequalities_graphically__diagram_03; asset_name: Boundary inclusion and shaded regions; caption: Dotted excludes equality; solid includes it; description: Side-by-side axes show the region y greater than x plus 1 above a dotted line, and y greater than or equal to x squared minus 2x minus 2 above a solid parabola. Both boundaries have labelled intercepts or turning points and clearly labelled shading.]
Dotted excludes equality

For $y>f(x)$ , shade above the graph; for $y<f(x)$ , shade below it. Test a point away from the boundary if unsure. These conventions are not in the formula booklet.

Worked example 11: a strict linear region

Question. Describe the graph of $y>x+1$ .

Plan. Locate the boundary line, exclude it, then test the origin.

\begin{aligned} y&=x+1 &&\text{boundary through }(-1, 0),\ (0, 1)\\ 0&>1\quad\text{is false} &&\text{test }(0, 0) \end{aligned}

Why this step? A test point decides the side independently of the drawing's scale.

Answer. Draw a dotted line and shade above it, excluding the line.

Check. $(0, 2)$ is shaded because $2>1$ ; $(0, 0)$ is not.

Change only $>$ to $\ge$ . The shaded side remains, and the boundary becomes solid.

Worked example 12: including a line

Question. Describe the graph of $y\ge x+1$ .

Plan. Use the same line and include equality.

\begin{aligned} y&=x+1 &&\text{same boundary}\\ 1&\ge1\quad\text{is true} &&\text{test boundary point }(0, 1) \end{aligned}

Why this step? The equality case adds the entire line, not a new shaded side.

Answer. Draw a solid line through $(-1, 0)$ , $(0, 1)$ , shading above.

Check. The origin still fails $0\ge1$ , so it stays unshaded.

A parabola uses precisely the same above-or-below reasoning. Locate its turning point to make the sketch meaningful.

Worked example 13: a strict quadratic region

Question. Describe the graph of $y>x^2-2x-2$ .

Plan. Complete the square for the boundary and exclude equality.

\begin{aligned} y&=(x-1)^2-3 &&\text{boundary has vertex }(1, -3)\\ 0&>-2\quad\text{is true} &&\text{origin is above the boundary} \end{aligned}

Why this step? “Above” compares vertical outputs at each input, even inside a curved region.

Answer. Draw a dotted upward parabola with vertex $(1, -3)$ , shading above it.

Check. $(1, -2)$ is shaded, since $-2>-3$ ; $(1, -4)$ is not.

Change only $>$ to $\ge$ , matching the right-hand panel of the diagram.

Worked example 14: including a parabola

Question. Describe the graph of $y\ge x^2-2x-2$ .

Plan. Keep the same parabola and shading, but include the boundary.

\begin{aligned} y&=(x-1)^2-3 &&\text{same boundary}\\ -3&\ge-3\quad\text{is true} &&\text{vertex is included} \end{aligned}

Why this step? A non-strict comparison includes every boundary point, so the whole curve is solid.

Answer. Solid upward parabola with vertex $(1, -3)$ , shading above.

Check. The origin still satisfies $0\ge-2$ .

One last surprising fact about the opening problem: $3/x<2$ has a broken solution set, although both sides look simple. The forbidden input $x=0$ divides its graph into two branches; algebra must respect that break too.

Watch out

Wrong: shading below because a parabola opens upwards. Its opening tells you the boundary's shape; the inequality sign decides the shaded side.

Use the same boundary-and-side decisions for a downward shaded region.

Quick Recap

Reverse the sign when multiplying or dividing by a negative number.
Find quadratic roots, inspect signs between and outside them, then include endpoints only when permitted.
Preserve denominator exclusions and use a multiplier with a known sign.
Compare two graphs by subtracting their outputs; use dotted or solid boundaries for planar regions.

For the opening inequality, multiply by $x^2>0$ with $x\ne0$ : $3/x<2$ gives $x<0$ or $x>3/2$ .

Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 2.5 - Linear and quadratic inequalities

Inequalities describe solution sets

Use roots to split a quadratic inequality

Reduce brackets and fractions safely

Interpret curve below line inequalities

Represent inequalities graphically

Quick Recap

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