Add Maths · 4 Equations, inequalities and graphs

4 Equations, inequalities and graphs

Before you start

The modulus $| a |$ is the size of $a$ without its sign, and $y = | f (x) |$ is $y = f (x)$ with the parts below the $x$ -axis reflected. The Functions lesson teaches this.
Solving quadratic equations by factorising or by the quadratic formula (it is on the List of formulas), and simplifying surds such as $\sqrt{28} = 2 \sqrt{7}$ .
Solving a quadratic inequality from a sketch: for $p < q$ , $(x - p) (x - q) < 0$ gives $p < x < q$ , and $(x - p) (x - q) > 0$ gives $x < p$ or $x > q$ .
The laws of indices, including $a^{\frac{1}{3}} = \sqrt[3]{a}$ and ${(a^{m})}^{n} = a^{m n}$ , and that $e^{x} = k$ gives $x = \ln k$ for $k > 0$ . The Logarithmic and exponential functions lesson covers $\ln$ in full.
Expanding three brackets, e.g. $(x + 3) (x - 1) (x - 2) = x^{3} - 7 x + 6$ .

Syllabus objectives covered (5)

Solve equations of the type • |ax + b| = c (c ⩾ 0) • |ax + b| = cx + d • |ax + b| = |cx + d| • |ax² + bx + c| = d using algebraic or graphical methods.
Solve graphically or algebraically inequalities of the type • k|ax + b| > c (c ⩾ 0) • k|ax + b| ⩽ c (c > 0) • k|ax + b| ⩽ |cx + d| where k > 0 • |ax + b| ⩽ cx + d • |ax² + bx + c| > d • |ax² + bx + c| ⩽ d
Use substitution to form and solve a quadratic equation in order to solve a related equation.
Sketch the graphs of cubic polynomials and their moduli, when given as a product of three linear factors.
Solve graphically cubic inequalities of the form • f(x) ⩾ d • f(x) > d • f(x) ⩽ d • f(x) < d where f(x) is a product of three linear factors and d is a constant.

Part 1 of 8

Equal or opposite: |ax + b| = c and |ax + b| = |cx + d|

$| A | = 7$ says the number $A$ is 7 units from zero. Exactly two numbers are: $7$ and $- 7$ . So one modulus equation splits into two ordinary ones.
$| A | = c ⟺ A = c or A = - c (c \geq 0)$
Solve $| 2 x - 5 | = 7$ . Write both cases down before you solve either, so the second one cannot be forgotten.
$2 x - 5 = 7 \Rightarrow x = 6 2 x - 5 = - 7 \Rightarrow x = - 1$
Check both in the original: $| 12 - 5 | = 7$ and $| - 2 - 5 | = | - 7 | = 7$ . With a constant $c \geq 0$ on the right, both cases always give genuine solutions, because the right-hand side can never be negative.
Graphically, $y = | 2 x - 5 |$ is a V with its point at $(\frac{5}{2}, 0)$ . The horizontal line $y = 7$ meets each arm once, and the $x$ -coordinates of those two meeting points are the solutions.
graph diagram — not drawn yet
Axes with x from -3 to 8 and y from -1 to 12, ticks at every unit. Solid V-shaped graph y = |2x - 5| with its point at (2.5, 0), meeting the y-axis at (0, 5), and passing through (-1, 7) and (6, 7). Dashed horizontal line y = 7 labelled 'y = 7'. Filled dots at (-1, 7) and (6, 7) with thin dotted vertical lines down to x = -1 and x = 6 on the x-axis. Label the left arm '2x - 5 = -7' and the right arm '2x - 5 = 7'.
Each arm of the V gives one case: the left arm is where $2 x - 5$ is negative, so it meets $y = 7$ where $2 x - 5 = - 7$ .
The picture explains the condition $c \geq 0$ . If $c = 0$ the line touches the point of the V, giving one solution, $x = \frac{5}{2}$ . If $c < 0$ the line is below the whole V and there are no solutions, because a modulus is never negative.
Now a modulus on each side. $| A | = | B |$ says $A$ and $B$ are the same distance from zero, so they are either equal or opposite.
$| A | = | B | ⟺ A = B or A = - B$
Solve $| 3 x - 1 | = | x + 5 |$ . Put the minus sign round the whole of one side, in a bracket, so that both of its terms change sign.
$3 x - 1 = x + 5 \Rightarrow x = 3 3 x - 1 = - (x + 5) \Rightarrow 4 x = - 4 \Rightarrow x = - 1$
A second valid route: both sides are never negative, so squaring them keeps the equation true and adds no false roots. ${(3 x - 1)}^{2} = {(x + 5)}^{2}$ becomes $8 x^{2} - 16 x - 24 = 0$ , which is $8 (x - 3) (x + 1) = 0$ . Same answers.
Check: $x = 3$ gives $| 8 | = | 8 |$ , and $x = - 1$ gives $| - 4 | = | 4 |$ . Both are true.

Part 2 of 8

When x is on the right: |ax + b| = cx + d

Now the right-hand side can change sign. The two cases still give you candidates, but a modulus can never equal a negative number. Any candidate that makes $c x + d$ negative is false and must be rejected.
Solve $| x - 4 | = 2 x + 1$ . Form both cases exactly as before.
$x - 4 = 2 x + 1 \Rightarrow x = - 5 x - 4 = - (2 x + 1) \Rightarrow 3 x = 3 \Rightarrow x = 1$
Test $x = 1$ in the original. Left: $| 1 - 4 | = 3$ . Right: $2 (1) + 1 = 3$ . It works.
Test $x = - 5$ . Left: $| - 5 - 4 | = 9$ . Right: $2 (- 5) + 1 = - 9$ . A modulus cannot equal $- 9$ , so reject $x = - 5$ . The only solution is $x = 1$ .
The graph shows where the false root came from. The line $y = 2 x + 1$ meets the V $y = | x - 4 |$ only once, at $(1, 3)$ . The case $x - 4 = 2 x + 1$ found where the line meets $y = x - 4$ , and at $x = - 5$ that line is below the axis, on the part the modulus reflected away.
graph diagram — not drawn yet
Axes with x from -6 to 6 and y from -10 to 10. Solid V-shaped graph y = |x - 4| with its point at (4, 0), meeting the y-axis at (0, 4). Dashed grey line y = x - 4 for x < 4, continuing below the x-axis through (0, -4) to (-6, -10): the part the modulus reflected. Solid straight line y = 2x + 1 through (0, 1). A filled dot at (1, 3) labelled '(1, 3): solution'. An open circle at (-5, -9), where the line y = 2x + 1 meets the dashed grey line, labelled 'x = -5: rejected'.
The rejected value is a real crossing, but with the dashed line $y = x - 4$ , which is not part of $y = | x - 4 |$ .
Write the rejection down with its reason: ' $x = - 5$ gives $2 x + 1 = - 9 < 0$ , so reject.' A final answer that lists both values is wrong, because $x = - 5$ does not satisfy the equation.
Squaring both sides is a valid way to find the candidates: ${(x - 4)}^{2} = {(2 x + 1)}^{2}$ gives $x^{2} + 4 x - 5 = 0$ , so $x = - 5$ or $x = 1$ . It produces the same false root, because squaring cannot tell $9$ from $- 9$ . You still have to check.

Part 3 of 8

The modulus of a quadratic: |ax² + bx + c| = d

The same split works with a quadratic inside: $| Q | = d$ means $Q = d$ or $Q = - d$ . Each case is a quadratic equation, so there can be up to four solutions.
Solve $| x^{2} - 4 x | = 3$ . First case: it does not factorise, so use the formula and simplify the surd.
$x^{2} - 4 x - 3 = 0 \Rightarrow x = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm 2 \sqrt{7}}{2} = 2 \pm \sqrt{7}$
Second case: this one factorises.
$x^{2} - 4 x = - 3 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow (x - 1) (x - 3) = 0 \Rightarrow x = 1 or x = 3$
All four are valid: the right-hand side is a non-negative constant, so neither case can give a false root. On a non-calculator paper, leave $2 \pm \sqrt{7}$ in surd form.
The graph tells you how many solutions to expect before you start. $x^{2} - 4 x = {(x - 2)}^{2} - 4$ has its minimum at $(2, - 4)$ . So $y = | x^{2} - 4 x |$ has an arch between $x = 0$ and $x = 4$ with its top at $(2, 4)$ .
graph diagram — not drawn yet
Axes with x from -1.5 to 5.5 and y from -5 to 7. Dashed grey parabola y = x^2 - 4x with vertex (2, -4) and roots (0, 0) and (4, 0). Solid bold graph y = |x^2 - 4x|: identical to the parabola for x <= 0 and x >= 4, and between x = 0 and x = 4 the reflected arch with its top at (2, 4); sharp corners at (0, 0) and (4, 0). Three horizontal lines: y = 5 (meets the bold graph twice, at x = -1 and x = 5), y = 4 (touches the top of the arch at (2, 4) and meets the outer branches at about x = -0.83 and x = 4.83, three points in all), and y = 3 (meets the bold graph four times, at about x = -0.65, x = 1, x = 3 and x = 4.65). Label each line with its number of meeting points: '2', '3', '4'. Label (2, -4) and (2, 4).
The number of solutions of $| x^{2} - 4 x | = d$ depends on where $y = d$ sits compared with the top of the arch, $(2, 4)$ .
Count the meetings with $y = d$ . Above the arch ( $d > 4$ ): two. Level with its top ( $d = 4$ ): three, because the line touches the top of the arch. Cutting through the arch ( $0 < d < 4$ ): four. At $d = 0$ : two, the roots $0$ and $4$ .
The algebra agrees. For $d > 4$ the second case is $x^{2} - 4 x + d = 0$ , whose discriminant $16 - 4 d$ is negative, so that case adds nothing.
So the height of the reflected turning point is the number that matters. Find it by completing the square whenever a question asks for the values of $d$ (or $k$ ) that give a stated number of solutions.

Part 4 of 8

Modulus inequalities against a constant

$| A | < c$ says $A$ is less than $c$ from zero, so $A$ lies between $- c$ and $c$ . $| A | > c$ says $A$ is further than $c$ from zero, so it is beyond $c$ on one side or the other.
$| A | < c ⟺ - c < A < c | A | > c ⟺ A < - c or A > c$
With a positive number $k$ in front, divide by $k$ first. Because $k > 0$ , the inequality sign stays the same.
Solve $3 | 2 x - 1 | \leq 15$ . Divide by 3, write the double inequality, then do the same thing to all three parts.
$| 2 x - 1 | \leq 5 \Rightarrow - 5 \leq 2 x - 1 \leq 5 \Rightarrow - 4 \leq 2 x \leq 6 \Rightarrow - 2 \leq x \leq 3$
Solve $2 | x + 3 | > 8$ . Divide by 2 to get $| x + 3 | > 4$ . This one splits into two separate pieces.
$x + 3 < - 4 or x + 3 > 4 \Rightarrow x < - 7 or x > 1$
On a graph, ' $\leq$ ' is where the V is on or below the horizontal line: one stretch, between the meeting points. ' $>$ ' is where the V is above the line: the two outer pieces. If you are unsure which, sketch the V.
With a quadratic inside, $| Q | \leq d$ means $- d \leq Q$ and $Q \leq d$ at the same time. That is two quadratic inequalities, and the answer is where both hold.
Solve $| x^{2} - 4 x | \leq 3$ . The critical values are the four solutions from the last section: $2 - \sqrt{7}$ , $1$ , $3$ and $2 + \sqrt{7}$ . On the graph, $y = | x^{2} - 4 x |$ is on or below $y = 3$ from $2 - \sqrt{7}$ to $1$ , rises above 3 over the arch, then is back below from $3$ to $2 + \sqrt{7}$ .
$2 - \sqrt{7} \leq x \leq 1 or 3 \leq x \leq 2 + \sqrt{7}$
Algebra gives the same: $x^{2} - 4 x \leq 3$ holds for $2 - \sqrt{7} \leq x \leq 2 + \sqrt{7}$ , and $x^{2} - 4 x \geq - 3$ holds for $x \leq 1$ or $x \geq 3$ . The answer is the overlap of the two.
$| x^{2} - 4 x | > 3$ is everything else, with the critical values left out.
$x < 2 - \sqrt{7} or 1 < x < 3 or x > 2 + \sqrt{7}$
Test one value in a region you chose. At $x = 2$ , $| 4 - 8 | = 4 > 3$ , so the middle region $1 < x < 3$ does belong to ' $> 3$ '.

Part 5 of 8

Modulus inequalities with x on both sides

When both sides contain $x$ , use two stages every time. Stage 1: solve the equation to find the critical values. Stage 2: decide which regions satisfy the inequality.
Solve $2 | x - 1 | \leq | x + 4 |$ . The critical values come from $2 | x - 1 | = | x + 4 |$ , which is 'equal or opposite'.
$2 (x - 1) = x + 4 \Rightarrow x = 6 2 (x - 1) = - (x + 4) \Rightarrow 3 x = - 2 \Rightarrow x = - \frac{2}{3}$
Now the regions. $y = 2 | x - 1 |$ is a V with its point at $(1, 0)$ and arms of gradient $\pm 2$ . $y = | x + 4 |$ is a V with its point at $(- 4, 0)$ and arms of gradient $\pm 1$ . The steeper V is underneath between the two crossings.
$- \frac{2}{3} \leq x \leq 6$
graph diagram — not drawn yet
Axes with x from -6 to 8 and y from -1 to 15. Solid V y = 2|x - 1| with its point at (1, 0), passing through (0, 2) and (6, 10). Second V in a different colour, y = |x + 4|, with its point at (-4, 0), passing through (0, 4) and (6, 10). The two graphs cross at (-2/3, 10/3) and (6, 10), both marked with filled dots and labelled. A bold segment on the x-axis from x = -2/3 to x = 6 with filled circles at both ends, labelled '2|x - 1| <= |x + 4|'.
Between the crossings the steeper V, $y = 2 | x - 1 |$ , lies under $y = | x + 4 |$ ; outside them it is above.
Check one value inside and one outside. $x = 0$ : $2 | - 1 | = 2$ and $| 4 | = 4$ , and $2 \leq 4$ is true. $x = 7$ : $2 | 6 | = 12$ and $| 11 | = 11$ , and $12 \leq 11$ is false. The region is right.
Because both sides are moduli, squaring is safe here too: both sides are never negative, and for non-negative numbers squaring keeps the order. The second worked example uses that route.
The type $| a x + b | \leq c x + d$ is different, because $c x + d$ can be negative. Do not square it. Use this instead, which is true for any $B$ , positive or not:
$| A | \leq B ⟺ - B \leq A \leq B$
Solve $| 2 x - 3 | \leq x + 1$ . Split the double inequality into its two halves and solve each.
$- (x + 1) \leq 2 x - 3 \Rightarrow 2 \leq 3 x \Rightarrow x \geq \frac{2}{3} 2 x - 3 \leq x + 1 \Rightarrow x \leq 4$
Both halves must hold, so $\frac{2}{3} \leq x \leq 4$ . On a graph, that is where the V $y = | 2 x - 3 |$ is on or below the line $y = x + 1$ .
If the line is steeper than the arms of the V, it meets the V only once, and the answer is a single inequality such as $x \geq 1$ . The $- B \leq A \leq B$ method handles that case with no change.

Part 6 of 8

Substitution: finding the hidden quadratic

Some equations are quadratics in disguise. The sign is two terms where one is the square of the other, such as $x$ and $\sqrt{x}$ , or $4^{x}$ and $2^{x}$ . Replace the one that is being squared by $u$ and you have an ordinary quadratic in $u$ .
Solve $x - 5 \sqrt{x} + 6 = 0$ . Since $x = {(\sqrt{x})}^{2}$ , let $u = \sqrt{x}$ .
$u^{2} - 5 u + 6 = 0 \Rightarrow (u - 2) (u - 3) = 0 \Rightarrow u = 2 or u = 3$
You have found $u$ , not $x$ . Go back: $\sqrt{x} = 2$ gives $x = 4$ , and $\sqrt{x} = 3$ gives $x = 9$ . Stopping at $u = 2, 3$ answers a different question.
Every value of $u$ must be possible for what it replaced. $\sqrt{x}$ is never negative, so a factor giving $u = - 2$ would be rejected. The same goes for $2^{x}$ and $e^{x}$ , which are always positive.
Fractional powers work the same way. In $x^{\frac{2}{3}} - 3 x^{\frac{1}{3}} - 4 = 0$ , $x^{\frac{2}{3}} = {(x^{\frac{1}{3}})}^{2}$ , so let $u = x^{\frac{1}{3}}$ .
$u^{2} - 3 u - 4 = 0 \Rightarrow (u - 4) (u + 1) = 0$
Cube each value to undo the cube root: $u = 4$ gives $x = 64$ , and $u = - 1$ gives $x = - 1$ . Here the negative value stays, because a cube root of a negative number exists: ${(- 1)}^{3} = - 1$ .
For exponentials, rewrite the larger power in terms of the smaller one first: $4^{x} = {(2^{2})}^{x} = {(2^{x})}^{2}$ , and $2^{x + 1} = 2 \times 2^{x}$ . So $4^{x} - 6 (2^{x}) + 8 = 0$ becomes $u^{2} - 6 u + 8 = 0$ with $u = 2^{x}$ .
$(u - 2) (u - 4) = 0 \Rightarrow 2^{x} = 2 or 2^{x} = 4 \Rightarrow x = 1 or x = 2$
A negative power is a reciprocal: $e^{- x} = \frac{1}{e^{x}}$ . Multiply the whole equation by $e^{x}$ , which is never zero, to clear it, and a quadratic in $e^{x}$ appears. The third worked example does this.
Logarithms fit the pattern too. In ${(\ln x)}^{2} - 3 \ln x + 2 = 0$ , let $u = \ln x$ : then $u = 1$ or $u = 2$ , so $x = e$ or $x = e^{2}$ . Here a negative $u$ would be allowed, since $\ln x$ is negative for $0 < x < 1$ .
Write the substitution as its own line, 'Let $u = 2^{x}$ ', followed by the quadratic in $u$ . That line is the method: it shows what you did even if an arithmetic slip comes later.

Part 7 of 8

Sketching a cubic from three factors, and its modulus

A cubic written as three linear factors, such as $y = (x + 3) (x - 1) (x - 2)$ , gives its $x$ -intercepts at once: each bracket is zero at one value of $x$ . Here they are $x = - 3$ , $x = 1$ and $x = 2$ .
The $y$ -intercept comes from $x = 0$ , which leaves the product of the constants: $(3) (- 1) (- 2) = 6$ . So the curve crosses the $y$ -axis at $(0, 6)$ .
Which way up comes from the $x^{3}$ term. Multiply the coefficients of $x$ in the three brackets: here $1 \times 1 \times 1 = 1$ . A positive $x^{3}$ term means $y$ is large and negative for large negative $x$ , and large and positive for large positive $x$ : the curve starts bottom left and ends top right.
If that product is negative, the curve is the other way up: it starts top left and ends bottom right. Watch for a bracket such as $(2 - x)$ : its $x$ -coefficient is $- 1$ , and that alone flips the curve.
Join the points with a smooth curve that crosses the axis at each root and turns once between each pair of neighbouring roots. The syllabus requires the points where the graph meets the axes to be clearly labelled. Turning points need differentiation to find exactly, so a sketch from factors shows them in roughly the right place.
graph diagram — not drawn yet
Two graphs side by side, each with x from -4 to 3 and y from -14 to 16. Left, titled 'y = (x + 3)(x - 1)(x - 2)': a smooth cubic rising from (-3.5, -12.4), crossing the x-axis at (-3, 0), reaching a local maximum at about (-1.53, 13.1), crossing the y-axis at (0, 6), crossing the x-axis at (1, 0), dipping to a local minimum at about (1.53, -1.13), crossing at (2, 0) and rising to (3, 12). Right, titled 'y = |(x + 3)(x - 1)(x - 2)|': the same curve with the part for x < -3 reflected above the x-axis (rising to the top left, through (-3.5, 12.4)) and the small dip between x = 1 and x = 2 reflected into a small hump with top at about (1.53, 1.13); sharp corners at (-3, 0), (1, 0) and (2, 0). On both graphs label (-3, 0), (1, 0), (2, 0) and (0, 6).
The modulus keeps the parts above the axis and flips the two parts below it; every crossing of the $x$ -axis becomes a sharp corner.
Check the sign in one region to be sure. At $x = 1.5$ : $(4.5) (0.5) (- 0.5) < 0$ , so the curve is below the axis between 1 and 2, as the sketch shows.
If two of the factors are the same, as in $y = {(x - 1)}^{2} (x + 2)$ , the curve touches the $x$ -axis at $x = 1$ and turns there instead of crossing, because ${(x - 1)}^{2}$ is never negative.
For $y = | f (x) |$ , keep everything on or above the $x$ -axis and reflect everything below it, exactly as in the Functions lesson. For $y = | (x + 3) (x - 1) (x - 2) |$ , two pieces flip: the tail for $x < - 3$ , which now rises to the top left, and the small dip between $x = 1$ and $x = 2$ , which becomes a small hump.
Label $(- 3, 0)$ , $(1, 0)$ , $(2, 0)$ and the $y$ -intercept, which for $| f (x) |$ is $(0, | f (0) |)$ . Here $| 6 | = 6$ , so it does not move. If $f (0)$ were negative, the intercept would move above the axis.

Part 8 of 8

Cubic inequalities from the graph

An inequality $f (x) > d$ asks where the curve $y = f (x)$ is above the horizontal line $y = d$ . So draw the line, find where it meets the curve, and read off the $x$ -intervals.
With $d = 0$ the line is the $x$ -axis and the roots are the critical values. For $f (x) = (x + 3) (x - 1) (x - 2)$ , the curve is above the axis between $- 3$ and $1$ , and again after $2$ .
$f (x) > 0 ⟺ - 3 < x < 1 or x > 2$
$f (x) \leq 0$ is where the curve is on or below the axis. The roots are included because the sign is ' $\leq$ '.
$f (x) \leq 0 ⟺ x \leq - 3 or 1 \leq x \leq 2$
For $d \neq 0$ , draw $y = d$ on an accurate graph. The line $y = 12$ meets this curve three times; reading the graph gives $x = - 2$ , $x = - 1$ and $x = 3$ .
graph diagram — not drawn yet
Axes with x from -4 to 4 and y from -14 to 26. Smooth cubic y = (x + 3)(x - 1)(x - 2) crossing the x-axis at (-3, 0), (1, 0) and (2, 0), with a local maximum at about (-1.53, 13.1), y-intercept (0, 6) and a local minimum at about (1.53, -1.13), rising to (3.5, 24.4). Dashed horizontal line y = 12 meeting the curve at (-2, 12), (-1, 12) and (3, 12), each marked with a filled dot and a dotted vertical line down to the x-axis. The parts of the curve on or above y = 12 (from x = -2 to x = -1 and from x = 3 onwards) drawn in a highlight colour, with matching bold segments on the x-axis: from -2 to -1 and from 3 to the right edge, filled circles at -2, -1 and 3.
$f (x) \geq 12$ is the highlighted part: the top of the hump between $- 2$ and $- 1$ , and everything from $3$ onwards.
When a reading looks like an exact value, verify it by substitution: $f (- 2) = (1) (- 3) (- 4) = 12$ , $f (- 1) = (2) (- 2) (- 3) = 12$ and $f (3) = (6) (2) (1) = 12$ .
Now read the regions. The curve is on or above $y = 12$ over the top of the hump, between $- 2$ and $- 1$ , and again from $3$ onwards.
$f (x) \geq 12 ⟺ - 2 \leq x \leq - 1 or x \geq 3$
$f (x) < 12$ is the rest, with the critical values left out: $x < - 2$ or $- 1 < x < 3$ .
If $y = d$ passes above the hump or below the dip, it meets the curve only once, and the answer is a single inequality on one side of that point. Count the meeting points on the graph before you write anything: that count decides the shape of the answer.

Worked examples

Example 1. (a) Solve $| x^{2} - x - 6 | = 6$ . (b) Find the set of values of $k$ for which $| x^{2} - x - 6 | = k$ has exactly four solutions.

Show the working

(a) Split into the two cases $x^{2} - x - 6 = 6$ and $x^{2} - x - 6 = - 6$ . The right-hand side is a non-negative constant, so every root you find will be valid.
First case: get zero on one side and factorise.
$x^{2} - x - 12 = 0 \Rightarrow (x - 4) (x + 3) = 0 \Rightarrow x = 4 or x = - 3$
Second case: the constants cancel, leaving a common factor of $x$ .
$x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0 or x = 1$
Check one from each case: $x = 4$ gives $16 - 4 - 6 = 6$ , and $x = 1$ gives $1 - 1 - 6 = - 6$ , whose modulus is 6.
(b) Find the height of the arch. Complete the square: $x^{2} - x - 6 = {(x - \frac{1}{2})}^{2} - \frac{25}{4}$ . The quadratic has its minimum $- \frac{25}{4}$ at $x = \frac{1}{2}$ , so the reflected arch has its top at $(\frac{1}{2}, \frac{25}{4})$ .
A horizontal line $y = k$ meets the graph four times when it cuts through the arch: above the $x$ -axis but below the top. Part (a) agrees: $6 < \frac{25}{4}$ , and there were four roots.
$0 < k < \frac{25}{4}$

Answer: (a) $x = - 3$ , $x = 0$ , $x = 1$ , $x = 4$ (b) $0 < k < \frac{25}{4}$

Example 2. Solve $3 | x + 1 | \leq | x - 5 |$ .

Show the working

Both sides are moduli, so neither is ever negative. For non-negative numbers, $p \leq q$ exactly when $p^{2} \leq q^{2}$ , so you may square both sides.
$9 {(x + 1)}^{2} \leq {(x - 5)}^{2}$
Expand both sides.
$9 x^{2} + 18 x + 9 \leq x^{2} - 10 x + 25$
Collect everything on one side, then divide by 4. Dividing by a positive number keeps the sign.
$8 x^{2} + 28 x - 16 \leq 0 \Rightarrow 2 x^{2} + 7 x - 4 \leq 0$
Factorise to find the critical values.
$(2 x - 1) (x + 4) \leq 0 critical values x = - 4, x = \frac{1}{2}$
$y = 2 x^{2} + 7 x - 4$ is a U, so it is on or below the axis between its roots, ends included.
$- 4 \leq x \leq \frac{1}{2}$
Check the ends and one inside value. $x = - 4$ : $3 | - 3 | = 9$ and $| - 9 | = 9$ . $x = \frac{1}{2}$ : $3 \times \frac{3}{2} = \frac{9}{2}$ and $| - \frac{9}{2} | = \frac{9}{2}$ . $x = 0$ : $3 \leq 5$ , true.
The case method finds the same critical values: $3 (x + 1) = x - 5$ gives $x = - 4$ , and $3 (x + 1) = - (x - 5)$ gives $x = \frac{1}{2}$ . You would then choose the region from a sketch of the two V shapes.

Answer: $- 4 \leq x \leq \frac{1}{2}$

Example 3. Solve $2 e^{x} = 5 + 3 e^{- x}$ , giving your answer in exact form.

Show the working

Spot the pair: $e^{x}$ and $e^{- x} = \frac{1}{e^{x}}$ . Multiply every term by $e^{x}$ , which is never zero, to clear the fraction.
$2 e^{2 x} = 5 e^{x} + 3$
Now $e^{2 x} = {(e^{x})}^{2}$ . Let $u = e^{x}$ and rearrange so one side is zero.
$2 u^{2} - 5 u - 3 = 0$
Factorise.
$(2 u + 1) (u - 3) = 0 \Rightarrow u = - \frac{1}{2} or u = 3$
Go back to $x$ . $e^{x} = - \frac{1}{2}$ is impossible, because $e^{x} > 0$ for every $x$ , so reject it.
$e^{x} = 3$ gives $x = \ln 3$ , because $\ln$ undoes $e^{x}$ .
Check: $2 e^{\ln 3} = 2 \times 3 = 6$ , and $5 + 3 e^{- \ln 3} = 5 + 3 \times \frac{1}{3} = 6$ .

Answer: $x = \ln 3$

Example 4. $f (x) = (x + 4) (x - 1) (3 - x)$ . (a) Sketch $y = f (x)$ and $y = | f (x) |$ , stating the coordinates of the points where each graph meets the axes. (b) Verify that $f (- 3) = f (- 1) = f (4) = - 24$ . (c) Hence solve $f (x) > - 24$ .

Show the working

(a) The roots are $x = - 4$ , $x = 1$ and $x = 3$ . The $y$ -intercept is $f (0) = (4) (- 1) (3) = - 12$ .
The $x$ -coefficients multiply to $1 \times 1 \times (- 1) = - 1$ , because of the $(3 - x)$ . The $x^{3}$ term is negative, so the curve starts top left and ends bottom right.
So it comes down through $(- 4, 0)$ , dips below the axis through $(0, - 12)$ , rises through $(1, 0)$ to a hump, and falls through $(3, 0)$ . Check one region: $f (2) = (6) (1) (1) = 6 > 0$ , so the hump between 1 and 3 is above the axis.
For $y = | f (x) |$ , reflect the two parts below the axis: the dip for $- 4 < x < 1$ , which becomes a hump, and the tail for $x > 3$ , which now rises to the top right. The $x$ -intercepts stay at $(- 4, 0)$ , $(1, 0)$ and $(3, 0)$ as sharp corners, and the $y$ -intercept moves to $(0, 12)$ .
(b) Substitute each value: $f (- 3) = (1) (- 4) (6) = - 24$ , $f (- 1) = (3) (- 2) (4) = - 24$ and $f (4) = (8) (3) (- 1) = - 24$ .
(c) Put $y = - 24$ on the sketch of $y = f (x)$ . It is below the axis, so it cuts the dip twice (at $- 3$ and $- 1$ ) and the falling tail once (at $4$ ). The dip really does go below $- 24$ between those points: $f (- 2) = (2) (- 3) (5) = - 30$ .
The curve is above the line to the left of $- 3$ , and from $- 1$ to $4$ (up out of the dip, over the hump and back down). It is below the line between $- 3$ and $- 1$ and after $4$ . The sign is strict, so the critical values are left out.
$x < - 3 or - 1 < x < 4$

Answer: (a) $y = f (x)$ : a cubic with negative $x^{3}$ term through $(- 4, 0)$ , $(0, - 12)$ , $(1, 0)$ and $(3, 0)$ . $y = | f (x) |$ : the parts for $- 4 < x < 1$ and $x > 3$ reflected, meeting the axes at $(- 4, 0)$ , $(1, 0)$ , $(3, 0)$ and $(0, 12)$ . (b) Each value is $- 24$ . (c) $x < - 3$ or $- 1 < x < 4$

Where people go wrong

Mistake: Giving both candidates for $| x - 4 | = 2 x + 1$ as answers without testing them.

Instead: Substitute each candidate into the original. $x = - 5$ makes the right-hand side $- 9$ , so reject it; the answer is $x = 1$ only.

A modulus is never negative, so the equation can only hold where $2 x + 1 \geq 0$ . The case split does not know that, so you have to apply it yourself.

Mistake: Writing $| x + 3 | > 4$ as $- 4 > x + 3 > 4$ , or as $- 4 < x + 3 < 4$ .

Instead: Write $x + 3 < - 4$ or $x + 3 > 4$ , so $x < - 7$ or $x > 1$ .

More than 4 from zero means beyond 4 on either side: two separate pieces joined by 'or'. The combined form $- c < A < c$ belongs to $| A | < c$ only.

Mistake: Squaring both sides of $| x - 2 | \leq 2 x - 1$ .

Instead: Use $- (2 x - 1) \leq x - 2 \leq 2 x - 1$ , which gives $x \geq 1$ .

Squaring gives $x^{2} \geq 1$ , which wrongly includes values such as $x = - 2$ , where the left side is $4$ and the right side is $- 5$ . Squaring an inequality is safe only when both sides are known to be non-negative, as with two moduli.

Mistake: Stopping at $u = 2$ and $u = 3$ for $x - 5 \sqrt{x} + 6 = 0$ , or keeping $e^{x} = - \frac{1}{2}$ as a solution.

Instead: Turn each $u$ back into $x$ ( $x = 4$ and $x = 9$ ), and reject any value of $u$ that the replaced expression cannot take, such as a negative value of $\sqrt{x}$ , $2^{x}$ or $e^{x}$ .

The question asks for $x$ . The letter $u$ is only a stepping stone, and it carries the restrictions of whatever it replaced.

Mistake: Sketching $y = (2 - x) (x + 1) (x + 3)$ as starting bottom left, or reflecting the whole cubic to draw $y = | f (x) |$ .

Instead: Multiply the $x$ -coefficients: $(- 1) (1) (1) = - 1$ , so this cubic starts top left. For the modulus, reflect only the parts below the $x$ -axis.

A bracket of the form $(a - x)$ has a negative $x$ -coefficient. The modulus changes only negative values; a positive value is already its own modulus.

Check yourself

Solve $| 4 x + 3 | = 9$ .
Answer
$4 x + 3 = 9$ gives $x = \frac{3}{2}$ ; $4 x + 3 = - 9$ gives $x = - 3$ . Both are valid, because the right-hand side is a non-negative constant.
Solve $| 2 x - 1 | = | x + 4 |$ .
Answer
$2 x - 1 = x + 4$ gives $x = 5$ ; $2 x - 1 = - (x + 4)$ gives $3 x = - 3$ , so $x = - 1$ . Check: $| 9 | = | 9 |$ and $| - 3 | = | 3 |$ .
Solve $| x + 5 | = 1 - x$ .
Answer
$x + 5 = 1 - x$ gives $x = - 2$ , and then $1 - x = 3 \geq 0$ , so it is valid: $| 3 | = 3$ . $x + 5 = - (1 - x) = x - 1$ gives $5 = - 1$ , which is impossible, so that case adds nothing. Answer: $x = - 2$ . (The line $y = 1 - x$ is parallel to the left arm of the V, so it meets the V only once.)
Solve $| x - 2 | \leq 2 x - 1$ .
Answer
Use $- (2 x - 1) \leq x - 2 \leq 2 x - 1$ . The left half gives $3 \leq 3 x$ , so $x \geq 1$ . The right half gives $x \geq - 1$ . Both must hold: $x \geq 1$ . Check $x = 1$ : $| - 1 | = 1 \leq 1$ .
Solve $| x^{2} - 5 | < 4$ .
Answer
$- 4 < x^{2} - 5 < 4$ , so $1 < x^{2} < 9$ . $x^{2} > 1$ gives $x < - 1$ or $x > 1$ ; $x^{2} < 9$ gives $- 3 < x < 3$ . Both hold for $- 3 < x < - 1$ or $1 < x < 3$ .
Solve $9^{x} - 4 (3^{x}) + 3 = 0$ .
Answer
$9^{x} = {(3^{x})}^{2}$ . Let $u = 3^{x}$ : $u^{2} - 4 u + 3 = 0$ , so $(u - 1) (u - 3) = 0$ and $u = 1$ or $u = 3$ . $3^{x} = 1$ gives $x = 0$ ; $3^{x} = 3$ gives $x = 1$ .
Sketch $y = (2 - x) (x + 1) (x + 3)$ , labelling where it meets the axes. Which parts are reflected when you sketch $y = | (2 - x) (x + 1) (x + 3) |$ ?
Answer
The $x$ -coefficients multiply to $- 1$ , so the curve starts top left and ends bottom right. It meets the axes at $(- 3, 0)$ , $(- 1, 0)$ , $(2, 0)$ and $(0, 6)$ . It is below the $x$ -axis for $- 3 < x < - 1$ and for $x > 2$ , so those two parts are reflected in the modulus graph. The $y$ -intercept stays at $(0, 6)$ .
Using your sketch from the previous question, solve $(2 - x) (x + 1) (x + 3) < 0$ .
Answer
The curve is below the $x$ -axis for $- 3 < x < - 1$ and for $x > 2$ . Answer: $- 3 < x < - 1$ or $x > 2$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Solve $| x + 2 | = 3 x - 2$ . 3 marks No calculator

Mark scheme

M1 forms x + 2 = 3x − 2 and x + 2 = −(3x − 2) oe, or squares both sides to reach 8x² − 16x = 0 oe
A1 x = 2 and x = 0 found, or x = 2 from the single case x + 2 = 3x − 2 after stating that 3x − 2 ≥ 0 makes x + 2 positive
A1 x = 0 rejected because 3(0) − 2 = −2 < 0 (or never found, by the justified single-case method); final answer x = 2 only

(b) Solve $2 | x - 2 | \leq | x + 1 |$ . 3 marks No calculator

Mark scheme

M1 squares both sides to 4(x − 2)² ≤ (x + 1)² oe, or forms 2(x − 2) = x + 1 and 2(x − 2) = −(x + 1)
A1 critical values 1 and 5
A1 1 ≤ x ≤ 5 oe, e.g. x ≥ 1 and x ≤ 5

Question 2. This question is about the graph of $y = | x^{2} - 6 x + 5 |$ .

(a) Solve $| x^{2} - 6 x + 5 | = 3$ , giving your answers in exact form. 3 marks No calculator

Mark scheme

M1 forms x² − 6x + 5 = 3 and x² − 6x + 5 = −3
A1 x = 3 ± √7, from x² − 6x + 2 = 0
A1 x = 2 and x = 4, from x² − 6x + 8 = 0

(b) Hence solve $| x^{2} - 6 x + 5 | < 3$ . 2 marks No calculator

Mark scheme

M1 selects the two intervals where the graph is below y = 3: from the smaller surd root to 2, and from 4 to the larger surd root; FT their four values from (a)
A1 3 − √7 < x < 2 or 4 < x < 3 + √7 oe

Question 3.

(a) Solve $2^{2 x + 1} - 9 (2^{x}) + 4 = 0$ . 4 marks No calculator

Mark scheme

B1 2^(2x+1) written as 2(2^x)², or as 2u² with u = 2^x
M1 solves 2u² − 9u + 4 = 0, e.g. (2u − 1)(u − 4) = 0
A1 u = ½ and u = 4
A1 x = −1 and x = 2

Question 4. $f (x) = (x + 2) (x - 1) (x - 7)$ .

(a) Sketch the graph of $y = | f (x) |$ , stating the coordinates of the points where it meets the axes. 3 marks No calculator

Mark scheme

B1 shape: a cubic with positive x³ term whose parts for x < −2 and 1 < x < 7 are reflected above the x-axis, with no part of the graph below the x-axis
B1 (−2, 0), (1, 0) and (7, 0) labelled
B1 (0, 14) labelled

(b) Verify that $f (- 3) = f (3) = f (6) = - 40$ . 1 mark No calculator

Mark scheme

B1 (−1)(−4)(−10) = −40, (5)(2)(−4) = −40 and (8)(5)(−1) = −40 all shown

(c) Hence, using a sketch of $y = f (x)$ , solve $f (x) \geq - 40$ . 2 marks No calculator

Mark scheme

M1 uses −3, 3 and 6 as critical values and selects where the curve y = f(x) is on or above the line y = −40
A1 −3 ≤ x ≤ 3 or x ≥ 6 oe

How this is examined

Solving a modulus inequality: critical values first, then the region: Solve an inequality such as $k | a x + b | \leq | c x + d |$ , $| a x + b | \leq c x + d$ or $| a x^{2} + b x + c | < d$ ; 3–5 marks. The critical values from the matching equation (by squaring when both sides are moduli, or by two cases), then the correct region written as one double inequality or as pieces joined by 'or', with $<$ or $\leq$ matching the question.
Spotting the substitution that turns an equation into a quadratic: Solve an equation in $2^{x}$ , $e^{x}$ , $\sqrt{x}$ , $x^{\frac{1}{3}}$ or $\ln x$ that hides a quadratic; 3–5 marks. A stated substitution such as 'let $u = 2^{x}$ ' with the quadratic in $u$ , both values of $u$ , any impossible value (a negative $2^{x}$ , $e^{x}$ or $\sqrt{x}$ ) rejected, and every answer given in terms of $x$ .
The modulus of a quadratic against a horizontal line: Solve $| a x^{2} + b x + c | = d$ , then use it to solve a related inequality or to find the values of $k$ that give a stated number of solutions; 3–5 marks across the parts. Both quadratics $Q = d$ and $Q = - d$ solved, roots kept exact as surds, and the number of solutions matched to the height of the reflected turning point found by completing the square.
Modulus equations: two cases, and checking when x is on the right: Solve $| a x + b | = c x + d$ or $| a x + b | = | c x + d |$ ; 3 marks. Both equations $A = B$ and $A = - B$ formed, with the minus sign on the whole bracket. When the right-hand side contains $x$ , each candidate substituted back and any value that makes it negative rejected, with the reason written.
Sketching a cubic and its modulus from three factors: Sketch $y = f (x)$ or $y = | f (x) |$ for $f (x)$ given as three linear factors, stating where it meets the axes; 2–3 marks. The right way up (from the sign of the product of the $x$ -coefficients), every $x$ -intercept and the $y$ -intercept labelled with coordinates, and for $| f (x) |$ only the parts below the axis reflected, with sharp corners on the $x$ -axis.
Reading a cubic inequality off its graph: Hence solve $f (x) \geq d$ or $f (x) < d$ , after a sketch or with the meeting points given; 2 marks. The $x$ -values where $y = d$ meets the curve, and the intervals where the curve is above or below the line, closed ends for $\leq$ or $\geq$ and open ends for $<$ or $>$ .

Words to know

Modulus: $| a |$ , the size of $a$ without its sign: its distance from zero. $| - 3 | = | 3 | = 3$ , and a modulus is never negative.
Case: One of the ordinary equations a modulus equation splits into: $| A | = B$ gives the cases $A = B$ and $A = - B$ .
Rejected root: A value produced by the method that does not satisfy the original equation, such as one that makes the right-hand side of $| A | = c x + d$ negative.
Critical values: The values of $x$ where the two sides of an inequality are equal. They are the ends of the regions in the solution.
Substitution: Replacing an expression such as $2^{x}$ , $e^{x}$ or $\sqrt{x}$ by a single letter, usually $u$ , so that the equation becomes a quadratic in $u$ .
Linear factor: A factor of the form $(a x + b)$ . Each linear factor of a cubic gives one $x$ -intercept, at $x = - \frac{b}{a}$ .

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