Linear and Quadratic Inequalities – OCR FSMQ Additional Maths

1 Manipulating Inequalities

AL7 is about manipulating inequalities. Almost everything you do to an equation you may also do to an inequality — with one crucial exception.

The one rule that is different

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

Operation	Safe?	Example
Add or subtract anything	Always safe	$x - 3 < 5 \Rightarrow x < 8$
Multiply/divide by a positive	Safe	$3x < 12 \Rightarrow x < 4$
Multiply/divide by a negative	Reverse the sign	$-x < 4 \Rightarrow x > -4$
Multiply by an unknown	Avoid — you do not know its sign	Rearrange instead
Square both sides	Avoid — squaring is not order-preserving	$-3 < 2$ but $9 > 4$

Worked Example 1 — The specification's own example

Solve $3 - 2x < 11$.

①Subtract $3$: $-2x < 8$

②Divide by $-2$ — and flip: $x > -4$

Avoid the flip entirely by moving the $x$ to the positive side instead: $3 - 2x < 11 \Rightarrow 3 < 11 + 2x \Rightarrow -8 < 2x \Rightarrow -4 < x$. Same answer, no sign rule to remember.

Worked Example 2 — A double inequality

Solve $-5 \leqslant 3x - 2 < 7$.

①Do the same thing to all three parts. Add $2$: $-3 \leqslant 3x < 9$

②Divide by $3$ (positive, so no flip): $-1 \leqslant x < 3$

Keep the strictness of each end. The left was $\leqslant$ and stays $\leqslant$; the right was $<$ and stays $<$.

2 Quadratic Inequalities

You cannot solve a quadratic inequality the way you solve a quadratic equation. $x^2 > 9$ does not give $x > 3$. The correct answer is $x > 3$ or $x < -3$.

The reliable three-step method (AL8)

1. Rearrange to $(\ldots) > 0$ or $(\ldots) < 0$
2. Solve the corresponding equation to find the critical values
3. Sketch the parabola and read off the region you need

The pattern worth memorising

For a positive $x^2$ coefficient with roots $\alpha < \beta$:
expression $< 0$ $\Rightarrow$ $\alpha < x < \beta$ (between — one interval)
expression $> 0$ $\Rightarrow$ $x < \alpha$ or $x > \beta$ (outside — two intervals)

Worked Example 3 — Less than zero

Solve $x^2 - 5x + 6 < 0$.

①Already in the right form. Solve $x^2 - 5x + 6 = 0$.

②$(x-2)(x-3) = 0$, so the critical values are $2$ and $3$.

③The $x^2$ coefficient is positive, so the curve dips below the axis between the roots.

④$2 < x < 3$

Test a point to be sure. At $x = 2.5$: $6.25 - 12.5 + 6 = -0.25 < 0$ ✓

Worked Example 4 — Greater than zero

Solve $x^2 + 2x - 15 \geqslant 0$.

①$x^2 + 2x - 15 = (x+5)(x-3)$, so critical values $-5$ and $3$.

②Positive $x^2$ coefficient, and we want $\geqslant 0$ — that is outside the roots.

③$x \leqslant -5$ or $x \geqslant 3$

Write "or", not "and". No number is both $\leqslant -5$ and $\geqslant 3$. Writing "$3 \leqslant x \leqslant -5$" is meaningless and scores zero.

Worked Example 5 — A negative $x^2$ coefficient

Solve $8 - 2x - x^2 > 0$.

①Multiply through by $-1$ and flip: $x^2 + 2x - 8 < 0$

②$(x+4)(x-2) = 0$, critical values $-4$ and $2$.

③Now the coefficient is positive and we want $< 0$: between the roots.

④$-4 < x < 2$

Turning the coefficient positive first means you only ever need to remember one picture. Just do not forget to flip the inequality when you multiply by $-1$.

Worked Example 6 — Rearranging first

Solve $x^2 < 4x + 5$.

①Move everything to the left: $x^2 - 4x - 5 < 0$

②$(x-5)(x+1) = 0$, critical values $5$ and $-1$.

③Between the roots: $-1 < x < 5$

Do not divide by $x$ to turn $x^2 < 4x + 5$ into $x < 4 + \tfrac5x$. You do not know the sign of $x$, so you do not know whether to flip.

3 Solving Graphically

AL8 asks for solutions algebraically and graphically. The graphical version answers the question "for which $x$ is one curve above the other?"

Sketch both sides of the inequality as separate graphs.
Find where they cross by solving the equation.
Read off the $x$-values where the required graph is higher.

Worked Example 7 — Reading a graph

The graphs of $y = x^2$ and $y = x + 2$ cross at $x = -1$ and $x = 2$. Use this to solve $x^2 < x + 2$.

①We want where the parabola is below the line.

②Between the crossing points the line is on top.

③$-1 < x < 2$

④Algebraic check: $x^2 - x - 2 < 0 \Rightarrow (x-2)(x+1) < 0 \Rightarrow -1 < x < 2$ ✓

4 Quick Reference

The flip

Multiply or divide by a negative $\Rightarrow$ reverse the sign.

Avoid the flip

Move the $x$ term to the positive side instead.

Never square

Squaring does not preserve order.

Never divide by $x$

Its sign is unknown.

Step 1

Get everything on one side, compared with $0$.

Step 2

Solve the equation for critical values.

Step 3

Sketch, then read the region.

$<0$, positive $a$

Between the roots: one interval.

$>0$, positive $a$

Outside the roots: two intervals, joined by "or".

Verify

Test one value from your answer region.

5 Practice Questions

Question 1

Solve $5x - 3 \geqslant 2x + 9$.

▶ Show solution

$3x \geqslant 12$

$x \geqslant 4$

Question 2

Solve $4 - 3x > 19$.

▶ Show solution

$-3x > 15$

Dividing by $-3$ reverses the sign: $x < -5$.

Question 3

Solve $-7 < 2x + 1 \leqslant 9$.

▶ Show solution

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

Divide by $2$: $-4 < x \leqslant 4$

Question 4

Solve $x^2 - 7x + 12 < 0$.

▶ Show solution

$(x-3)(x-4) = 0$, critical values $3$ and $4$.

Positive $x^2$ coefficient and $< 0$, so between the roots:

$3 < x < 4$

Question 5

Solve $x^2 \geqslant 16$.

▶ Show solution

$x^2 - 16 \geqslant 0$, so $(x-4)(x+4) \geqslant 0$.

Critical values $\pm 4$; we want outside:

$x \leqslant -4$ or $x \geqslant 4$

Question 6

Solve $2x^2 + 5x - 3 \leqslant 0$.

▶ Show solution

$2x^2 + 5x - 3 = (2x - 1)(x + 3)$

Critical values $x = \tfrac12$ and $x = -3$.

Positive coefficient, $\leqslant 0$, so between (inclusive):

$-3 \leqslant x \leqslant \dfrac{1}{2}$

Question 7

Solve $12 - x - x^2 \geqslant 0$.

▶ Show solution

Multiply by $-1$ and flip: $x^2 + x - 12 \leqslant 0$

$(x+4)(x-3) \leqslant 0$, critical values $-4$ and $3$.

Between the roots: $-4 \leqslant x \leqslant 3$

Question 8

Solve $x(x - 1) > 6$.

▶ Show solution

Expand and rearrange: $x^2 - x - 6 > 0$

$(x-3)(x+2) > 0$, critical values $3$ and $-2$.

Outside the roots: $x < -2$ or $x > 3$

Check $x = 4$: $4 \times 3 = 12 > 6$ ✓; $x = 1$: $1 \times 0 = 0 \not> 6$ ✓

Question 9

Find the values of $x$ for which $x^2 + 4x + 7$ is less than $4$.

▶ Show solution

$x^2 + 4x + 7 < 4 \;\Rightarrow\; x^2 + 4x + 3 < 0$

$(x+1)(x+3) < 0$, critical values $-1$ and $-3$.

$-3 < x < -1$

Question 10

A rectangular pen is made with $40$ m of fencing, using a wall as one long side. The width is $x$ m.

(a) Show that the area is $A = x(40 - 2x)$. (b) Find the values of $x$ for which the area is at least $150$ m². (c) State the range of $x$ for which the pen exists at all, and combine this with (b). (d) Find the greatest possible area.

▶ Show solution

(a) Two widths of $x$ use $2x$ m of fencing, leaving $40 - 2x$ m for the side opposite the wall.

$A = x(40 - 2x)$

(b) $x(40-2x) \geqslant 150$

$40x - 2x^2 \geqslant 150 \;\Rightarrow\; -2x^2 + 40x - 150 \geqslant 0$

Divide by $-2$ and flip: $x^2 - 20x + 75 \leqslant 0$

$(x-5)(x-15) \leqslant 0$, so $5 \leqslant x \leqslant 15$.

(c) For the pen to exist, $x > 0$ and $40 - 2x > 0$, i.e. $0 < x < 20$.

The interval $5 \leqslant x \leqslant 15$ sits entirely inside $0 < x < 20$, so the answer to (b) stands unchanged: $5 \leqslant x \leqslant 15$.

(d) Complete the square: $A = -2\left(x^2 - 20x\right) = -2\left[(x-10)^2 - 100\right] = 200 - 2(x-10)^2$.

The greatest area is $200$ m², when $x = 10$ m (so the pen is $10$ m by $20$ m).

Note $x = 10$ lies in the middle of the interval from (b), as you would expect — the maximum is as far as possible from both ends where $A$ drops to $150$.

Linear and Quadratic Inequalities (AL7–AL8) · OCR FSMQ Additional Maths · Created with MathJax