Linear and Quadratic Inequalities – A Level Maths Year 1

1 Linear Inequalities

An inequality is solved almost exactly like an equation. You may add or subtract anything on both sides, and multiply or divide both sides by a positive number. The one new rule is about negatives.

The golden rule

Multiplying or dividing both sides by a negative number reverses the inequality sign
$-2x < 6 \;\Rightarrow\; x > -3$

Worked Example 1 — Unknowns on both sides

Solve $5 - 3x < 2x + 20$.

①Subtract $2x$ and $5$ from both sides: $-5x < 15$

②Divide by $-5$ and reverse the sign: $x > -3$

To avoid dividing by a negative at all, collect the $x$ terms on the side where they are positive: $5 - 20 < 2x + 3x$ gives $-15 < 5x$, so $-3 < x$. Same answer.

Worked Example 2 — A double inequality with a fraction

Solve $-3 \leqslant \dfrac{2x + 1}{3} < 3$.

①Multiply all three parts by $3$: $-9 \leqslant 2x + 1 < 9$

②Subtract $1$ throughout: $-10 \leqslant 2x < 8$

③Divide by $2$: $-5 \leqslant x < 4$

2 Writing the Answer: Inequality and Set Notation

A Level questions may ask for the answer in set notation. You need to be able to read and write all of these forms.

Inequality	Set notation
$x > -3$	$\{x : x > -3\}$
$x \leqslant 2$	$\{x : x \leqslant 2\}$
$-5 \leqslant x < 4$	$\{x : -5 \leqslant x < 4\}$ or $\{x : x \geqslant -5\} \cap \{x : x < 4\}$
$x < -3$ or $x > 4$	$\{x : x < -3\} \cup \{x : x > 4\}$

Reading the symbols. $\{x : \ldots\}$ means "the set of all $x$ such that …". $\cap$ (intersection) means and — both conditions hold. $\cup$ (union) means or — at least one holds.

Worked Example 3 — Two conditions at once

Find the set of values of $x$ for which $3x - 2 > 4$ and $2x + 1 \leqslant 15$.

①$3x > 6 \Rightarrow x > 2$

②$2x \leqslant 14 \Rightarrow x \leqslant 7$

③Both must hold: $2 < x \leqslant 7$, i.e. $\{x : x > 2\} \cap \{x : x \leqslant 7\} = \{x : 2 < x \leqslant 7\}$.

Never write $-3 > x > 4$ for "$x < -3$ or $x > 4$". A chained inequality means and, and no number is both less than $-3$ and greater than $4$. Two separate pieces need "or", or a union $\cup$.

3 Quadratic Inequalities

Three steps, every time

① Rearrange to $\mathrm{f}(x) > 0$ (or $<$, $\geqslant$, $\leqslant$) ② Solve $\mathrm{f}(x) = 0$ for the critical values ③ Sketch $y = \mathrm{f}(x)$ and read off where it is above or below the axis

Worked Example 4 — Outside the roots

Solve $x^2 - x - 12 > 0$, giving your answer in set notation.

①Critical values: $(x - 4)(x + 3) = 0$, so $x = -3$ and $x = 4$.

②$y = x^2 - x - 12$ is $\cup$-shaped, crossing the axis at $-3$ and $4$.

③We need $y > 0$: the parts of the curve above the axis, which are outside the roots.

④$x < -3$ or $x > 4$, i.e. $\{x : x < -3\} \cup \{x : x > 4\}$.

Worked Example 5 — Negative $x^2$ coefficient

Solve $6 + x - 2x^2 \geqslant 0$.

①Multiply by $-1$ (reversing the sign) to get a positive $x^2$ term: $2x^2 - x - 6 \leqslant 0$.

②$(2x + 3)(x - 2) = 0$ gives critical values $x = -\tfrac32$ and $x = 2$.

③The $\cup$-shaped curve is on or below the axis between (and at) the roots.

④$-\tfrac32 \leqslant x \leqslant 2$

You could equally sketch the $\cap$-shaped $y = 6 + x - 2x^2$ directly and look for where it is above the axis. Either way, the sketch does the thinking.

Worked Example 6 — Critical values that are surds

Solve $x^2 + 4x - 1 < 0$, giving exact answers.

①Complete the square: $(x + 2)^2 - 5 = 0$, so $x = -2 \pm \sqrt{5}$.

②$\cup$-shaped curve, below the axis between the roots.

③$-2 - \sqrt{5} < x < -2 + \sqrt{5}$

Worked Example 7 — An inequality with $x$ in the denominator

Solve $\dfrac{3}{x} > 1$, where $x \neq 0$.

①We cannot multiply by $x$ — we don't know its sign. Multiply by $x^2$ instead, which is certainly positive.

②$3x > x^2 \;\Rightarrow\; x^2 - 3x < 0 \;\Rightarrow\; x(x - 3) < 0$

③Critical values $0$ and $3$; the $\cup$-shaped curve is negative between them.

④$0 < x < 3$

Multiplying by $x$ would give $3 > x$, which wrongly includes every negative number. Try $x = -1$: $\tfrac{3}{-1} = -3$, which is not greater than $1$.

Worked Example 8 — From a discriminant condition

Find the set of values of $k$ for which $x^2 + kx + (k + 3) = 0$ has two distinct real roots.

①Two distinct real roots: $k^2 - 4(k + 3) > 0 \;\Rightarrow\; k^2 - 4k - 12 > 0$.

②$(k - 6)(k + 2) > 0$: critical values $-2$ and $6$.

③Outside the roots: $\{k : k < -2\} \cup \{k : k > 6\}$.

This is where most A Level quadratic inequalities come from — see Quadratic Functions and Simultaneous Equations.

4 Interpreting Inequalities Graphically

$\mathrm{f}(x) > \mathrm{g}(x)$ asks: for which $x$ is the graph of $y = \mathrm{f}(x)$ above the graph of $y = \mathrm{g}(x)$? The critical values are the $x$-coordinates of the points where the graphs cross.

Worked Example 9 — Curve above a line

Find the values of $x$ for which the curve $y = x^2$ lies above the line $y = 2x + 3$.

①Need $x^2 > 2x + 3$, i.e. $x^2 - 2x - 3 > 0$.

②$(x - 3)(x + 1) > 0$: the graphs cross at $x = -1$ and $x = 3$.

③The parabola is above the line outside these: $x < -1$ or $x > 3$.

Regions in two variables. An inequality in $x$ and $y$, such as $y < 4 - x^2$, describes a region of the plane. Draw the boundary curve, then decide which side is wanted.

Drawing regions

Solid boundary for $\leqslant$ or $\geqslant$ (points on it are included)
Dashed boundary for $<$ or $>$ (points on it are not included)
$y > \mathrm{f}(x)$ is above the curve; $y < \mathrm{f}(x)$ is below it

Worked Example 10 — A region between a line and a curve

Sketch the region $R$ satisfying $y < 4 - x^2$ and $y \geqslant x + 2$.

①Boundaries meet where $4 - x^2 = x + 2$: $x^2 + x - 2 = 0$, so $(x + 2)(x - 1) = 0$, giving $(-2,\,0)$ and $(1,\,3)$.

②$y < 4 - x^2$: below the parabola, which is drawn dashed.

③$y \geqslant x + 2$: on or above the line, which is drawn solid.

④$R$ is the region enclosed between them, for $-2 < x < 1$.

Test a point. $(0,\,3)$: is $3 < 4$? Yes. Is $3 \geqslant 2$? Yes. So $(0, 3)$ is in $R$ ✓. Conventions differ between questions — some ask you to shade the region wanted, others the region not wanted. Always read the instruction and label the region.

5 Quick Reference

Negative multiplier

Multiplying or dividing by a negative reverses the sign.

Quadratic method

Rearrange to $\mathrm{f}(x) \gtrless 0$, find critical values, sketch.

$\cup$-shape, $> 0$

Outside the roots: two pieces.

$\cup$-shape, $< 0$

Between the roots: one interval.

Set notation

$\{x : \ldots\}$; $\cap$ = and, $\cup$ = or.

Intervals

$[\;]$ includes the end point, $(\;)$ excludes it.

$x$ in denominator

Multiply by $x^2$, not $x$.

Graphically

$\mathrm{f}(x) > \mathrm{g}(x)$: where $\mathrm{f}$'s graph is above $\mathrm{g}$'s.

Boundaries

Solid for $\leqslant, \geqslant$; dashed for $<, >$.

Don't divide by $x$

$x^2 < 6x$: rearrange to $x(x-6) < 0$ instead.

6 Practice Questions

Question 1

Solve $7 - 2x \geqslant 3x - 8$, giving your answer in set notation.

▶ Show solution

$7 + 8 \geqslant 3x + 2x \;\Rightarrow\; 15 \geqslant 5x \;\Rightarrow\; x \leqslant 3$

$\{x : x \leqslant 3\}$

Question 2

Solve $-1 < \dfrac{3 - 2x}{5} \leqslant 3$.

▶ Show solution

Multiply by $5$: $-5 < 3 - 2x \leqslant 15$

Subtract $3$: $-8 < -2x \leqslant 12$

Divide by $-2$, reversing both signs: $4 > x \geqslant -6$

$-6 \leqslant x < 4$

Question 3

Solve $x^2 - 5x + 4 \leqslant 0$.

▶ Show solution

$(x - 1)(x - 4) = 0$ gives critical values $1$ and $4$.

$\cup$-shaped curve, on or below the axis between the roots: $1 \leqslant x \leqslant 4$.

Question 4

Solve $2x^2 + 5x - 3 > 0$, giving your answer in set notation.

▶ Show solution

$(2x - 1)(x + 3) = 0$ gives critical values $x = \tfrac12$ and $x = -3$.

$\cup$-shaped, above the axis outside the roots:

$\{x : x < -3\} \cup \left\{x : x > \tfrac12\right\}$

Question 5

Solve $x^2 < 6x$.

▶ Show solution

Do not divide by $x$ (its sign is unknown). Rearrange: $x^2 - 6x < 0$, so $x(x - 6) < 0$.

Critical values $0$ and $6$; negative between them.

$0 < x < 6$

Question 6

Find the set of values of $x$ for which both $2x - 5 < 7$ and $x^2 - 3x - 10 > 0$.

▶ Show solution

First: $2x < 12 \Rightarrow x < 6$.

Second: $(x - 5)(x + 2) > 0 \Rightarrow x < -2$ or $x > 5$.

A number line shows where both hold: everything below $-2$, and the gap between $5$ and $6$.

$\{x : x < -2\} \cup \{x : 5 < x < 6\}$

Question 7

Solve $x^2 + 2x - 5 \geqslant 0$, giving your answer in exact form.

▶ Show solution

$(x + 1)^2 - 6 = 0 \Rightarrow x = -1 \pm \sqrt{6}$.

$\cup$-shaped, on or above the axis outside the roots:

$x \leqslant -1 - \sqrt{6}$ or $x \geqslant -1 + \sqrt{6}$.

Question 8

Find the set of values of $k$ for which $x^2 + (k - 1)x + 4 = 0$ has no real roots.

▶ Show solution

No real roots: $(k - 1)^2 - 16 < 0$.

$(k - 1)^2 < 16 \Rightarrow -4 < k - 1 < 4$.

$-3 < k < 5$

(Alternatively $k^2 - 2k - 15 < 0 \Rightarrow (k - 5)(k + 3) < 0$, the same answer.)

Question 9

The region $R$ is defined by $y \geqslant x^2 - 3x + 2$ and $y < 4 - x$.

(a) State which boundary should be drawn solid and which dashed. (b) Determine whether each of the points $(1,\, 2)$ and $(3,\, 1)$ lies in $R$.

▶ Show solution

(a) $y = x^2 - 3x + 2$ is solid ($\geqslant$); $y = 4 - x$ is dashed ($<$).

(b) $(1, 2)$: $1 - 3 + 2 = 0$ and $2 \geqslant 0$ ✓; $4 - 1 = 3$ and $2 < 3$ ✓. So $(1, 2)$ is in $R$.

$(3, 1)$: $9 - 9 + 2 = 2$, and $1 \geqslant 2$ is false. So $(3, 1)$ is not in $R$ (it lies below the parabola).

Question 10

A farmer has $40$ m of fencing to make a rectangular pen. One side of the pen is $x$ m long.

(a) Show that the area of the pen is $A = x(20 - x)$ m$^2$, and state the values of $x$ that make sense. (b) The area must be at least $96$ m$^2$. Find the possible values of $x$. (c) By completing the square, show that the area can never exceed $100$ m$^2$. (d) Find the values of $x$ for which the area exceeds $99$ m$^2$.

▶ Show solution

(a) Perimeter $2x + 2w = 40$, so the other side is $w = 20 - x$, and $A = x(20 - x)$. Both sides must be positive, so $0 < x < 20$.

(b) $x(20 - x) \geqslant 96 \;\Rightarrow\; 20x - x^2 - 96 \geqslant 0 \;\Rightarrow\; x^2 - 20x + 96 \leqslant 0$

$(x - 8)(x - 12) \leqslant 0$, so $8 \leqslant x \leqslant 12$ (which lies within $0 < x < 20$).

(c) $A = 20x - x^2 = -\left(x^2 - 20x\right) = -\left[(x - 10)^2 - 100\right] = 100 - (x - 10)^2$

$(x - 10)^2 \geqslant 0$, so $A \leqslant 100$, with the maximum when $x = 10$ (a square pen).

(d) $100 - (x - 10)^2 > 99 \;\Rightarrow\; (x - 10)^2 < 1 \;\Rightarrow\; -1 < x - 10 < 1$

$9 < x < 11$

Linear and Quadratic Inequalities · A Level Maths Year 1 · Created with MathJax