Modulus Equations
Solving \(|ax+b| = |cx+d|\) — algebraic & graphical methods · §4.1
The modulus — recap
The modulus (absolute value) of x is the distance from zero — always non-negative.
\(|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}\)
Key property: \(|A| = |B| \iff A = B \text{ or } A = -B\). This gives the two cases to solve.
Two methods — when to use each
Algebraic
Use when: the expressions inside the moduli are linear and you want exact solutions quickly.
Graphical
Use when: asked to solve "graphically", or when the equation involves quadratic/cubic expressions. Helps you see how many solutions there are.
Algebraic method — solve \(|3x-2| = |x+4|\)
Case 1
\(3x-2 = x+4\)
\(2x = 6 \Rightarrow x = 3\)
\(2x = 6 \Rightarrow x = 3\)
Case 2
\(3x-2 = -(x+4)\)
\(3x-2 = -x-4\)
\(4x = -2 \Rightarrow x = -\tfrac{1}{2}\)
\(3x-2 = -x-4\)
\(4x = -2 \Rightarrow x = -\tfrac{1}{2}\)
Check
Substitute both back: \(x=3\): \(|7|=|7|\) ✓ \(x=-\tfrac{1}{2}\): \(|-\tfrac{7}{2}|=|\tfrac{7}{2}|\) ✓
✓
\(x = 3\) or \(x = -\tfrac{1}{2}\)
Always check your answers! Substituting back into the original modulus equation confirms validity — extraneous solutions can appear if you square both sides as an alternative method.
Graphical method — sketch and read off intersections
- Sketch \(y = |ax+b|\): draw \(y=ax+b\), reflect any part below x-axis upward
- Sketch \(y = |cx+d|\) on the same axes
- Read off the x-coordinates of intersection points
- If exact values needed, set up algebraic cases at the intersection regions
Number of solutions: Two V-shaped modulus graphs can intersect 0, 1, or 2 times. The sketch immediately tells you which case you're in.
Interactive — Modulus Equation Explorer
Solve \(|ax + b| = |cx + d|\) — adjust parameters and see the graph
Left side: |ax + b|
Right side: |cx + d|
—
Modulus Inequalities
Solving \(|f(x)| < k\), \(|f(x)| > k\), and \(|f(x)| < |g(x)|\) · §4.2
Core rules
\(|f(x)| < k \iff -k < f(x) < k\)
\(|f(x)| > k \iff f(x) > k \text{ or } f(x) < -k\)
These only apply when k > 0. If k ≤ 0, \(|f(x)|<k\) has no solutions; \(|f(x)|>k\) is always true.
Remember: "less than" → bounded interval (AND). "Greater than" → unbounded (OR). Getting these switched is the most common mistake.
Solve \(|2x - 3| < 5\)
Rule
\(|f(x)| < k \Rightarrow -k < f(x) < k\)
Apply
\(-5 < 2x - 3 < 5\)
+3
\(-2 < 2x < 8\)
÷2
\(-1 < x < 4\)
Solve \(|3x + 1| \geq 7\)
Rule
\(|f(x)| \geq k \Rightarrow f(x) \geq k \text{ or } f(x) \leq -k\)
Case 1
\(3x+1 \geq 7 \Rightarrow x \geq 2\)
Case 2
\(3x+1 \leq -7 \Rightarrow x \leq -\tfrac{8}{3}\)
✓
\(x \leq -\tfrac{8}{3}\) or \(x \geq 2\)
Solving \(|ax+b| < |cx+d|\) — squaring both sides
Since both sides are non-negative, squaring preserves the inequality direction.
\(|f(x)| < |g(x)| \iff [f(x)]^2 < [g(x)]^2\)
This converts the modulus inequality into a standard quadratic/polynomial inequality which you solve by factorising and sketching.
Only valid when both sides are non-negative — which is guaranteed when both sides are moduli.
Solve \(|x - 1| < |2x + 3|\)
Square
\((x-1)^2 < (2x+3)^2\)
Expand
\(x^2-2x+1 < 4x^2+12x+9\)
Rearrange
\(0 < 3x^2+14x+8\)
Factorise
\(0 < (3x+2)(x+4)\)
Sketch
Parabola ∪ shape, roots at \(x=-4\) and \(x=-\tfrac{2}{3}\)
✓
\(x < -4\) or \(x > -\tfrac{2}{3}\)
Graphical method for inequalities
|f(x)| < k
Find where graph of y = |f(x)| lies below the horizontal line y = k. Solution: the x-interval between the two intersection points.
|f(x)| > k
Find where graph lies above y = k. Solution: two unbounded regions outside the intersection points.
|f(x)| < |g(x)|
Find where graph of y = |f(x)| lies below y = |g(x)|. Mark intersection x-values, test a point in each region.
Cubic Graphs & Their Moduli
Sketching y = f(x), y = |f(x)|, and y = f(|x|) · §4.3
Sketching a cubic — key features to find
â‘ Roots (x-intercepts)
Factorise fully. Mark where the curve crosses the x-axis. Repeated roots = curve touches & bounces. Degree 3 → at most 3 real roots.
â‘¡ y-intercept
Substitute x = 0. For \(f(x) = a(x-p)(x-q)(x-r)\), the y-intercept is \(f(0) = a(-p)(-q)(-r)\).
â‘¢ End behaviour
Positive leading coeff: bottom-left to top-right. Negative: top-left to bottom-right. The cubic never turns back on itself.
Transforming to y = |f(x)|
- Sketch y = f(x) first, marking all roots and the y-intercept
- Any part of the curve above the x-axis stays unchanged
- Any part below the x-axis gets reflected upward (flip across x-axis)
- The resulting graph is entirely at or above the x-axis
Result: Roots where curve was below become local minima that touch the x-axis (they don't cross). Former troughs become peaks.
Transforming to y = f(|x|)
- Keep the part of y = f(x) for \(x \geq 0\) unchanged
- Reflect that right-hand portion in the y-axis to create the left-hand side
- Discard the original left-hand portion of the graph
Result: The graph is symmetric about the y-axis. It only uses the right-hand half of the original function.
Don't confuse: \(y=|f(x)|\) reflects below the x-axis upward. \(y=f(|x|)\) reflects the right side into the left side. Completely different transformations.
Interactive — Cubic Modulus Explorer
\(f(x) = (x-p)(x-q)(x-r)\) — choose roots and transformation
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Cubic Inequalities — Graphically
Using the sketch of a cubic to read off solution sets · §4.4
The graphical approach — why and how
Cubic inequalities cannot be solved by the simple "critical values and test" method that works for quadratics, because the shape is more complex. The graphical method is reliable and required by the syllabus.
- Rearrange so all terms are on one side: \(f(x) > 0\) or \(f(x) < 0\)
- Fully factorise \(f(x)\) — use the Factor Theorem to find roots
- Sketch the cubic \(y = f(x)\), marking all roots and y-intercept
- Read off where the curve is above (for > 0) or below (for < 0) the x-axis
- Write the solution as a set of intervals
Solve \((x-1)(x+2)(x-3) > 0\)
Roots
x = 1, −2, 3 — three distinct real roots
Leading
Positive leading coeff → starts bottom-left, ends top-right
Sketch
Cubic crosses x-axis at −2, 1, 3 in that order. Curve is above the x-axis for \(-2 < x < 1\) and \(x > 3\)
✓
\(-2 < x < 1\) or \(x > 3\)
Solving a cubic with a line — \(f(x) > g(x)\)
- Rearrange: \(f(x) - g(x) > 0\) and let \(h(x) = f(x) - g(x)\)
- Find roots of \(h(x) = 0\) using Factor Theorem
- Sketch \(y = h(x)\) and find where it is positive
Alternative: Sketch both \(y=f(x)\) and \(y=g(x)\) on the same axes. Find where the cubic is above the line — that's your solution set.
Sign analysis for quick working
For a factorised cubic \((x-a)(x-b)(x-c)\), use a sign table to determine the sign in each interval without sketching:
| Interval | (x−a) | (x−b) | (x−c) | Product |
| x < a | − | − | − | − |
| a < x < b | + | − | − | + |
| b < x < c | + | + | − | − |
| x > c | + | + | + | + |
(a < b < c, positive leading coefficient)
Equations Reducible to Quadratic
Substitution method · Disguised quadratics · §4.5
The core idea
Many equations that look complicated are secretly quadratics in disguise. The key is spotting that one expression is the square of another, then making a substitution.
General form to spot
If equation has the form:
\[a[f(x)]^2 + b[f(x)] + c = 0\]
let \(u = f(x)\) → \(au^2 + bu + c = 0\)
\[a[f(x)]^2 + b[f(x)] + c = 0\]
let \(u = f(x)\) → \(au^2 + bu + c = 0\)
Always substitute back! After solving for u, substitute back into u = f(x) to find x. Some values of u may give no real x — state this clearly.
Common disguised forms
Type 1
\(x^4 - 5x^2 + 4 = 0\)
Let \(u = x^2\) → \(u^2-5u+4=0\)
\(x^4 - 5x^2 + 4 = 0\)
Let \(u = x^2\) → \(u^2-5u+4=0\)
Type 2
\(e^{2x} - 3e^x + 2 = 0\)
Let \(u = e^x\) → \(u^2-3u+2=0\)
\(e^{2x} - 3e^x + 2 = 0\)
Let \(u = e^x\) → \(u^2-3u+2=0\)
Type 3
\(\sin^2 x - \sin x - 2 = 0\)
Let \(u = \sin x\) → \(u^2-u-2=0\)
\(\sin^2 x - \sin x - 2 = 0\)
Let \(u = \sin x\) → \(u^2-u-2=0\)
Type 4
\(2x - 7\sqrt{x} + 3 = 0\)
Let \(u = \sqrt{x}\) → \(2u^2-7u+3=0\)
\(2x - 7\sqrt{x} + 3 = 0\)
Let \(u = \sqrt{x}\) → \(2u^2-7u+3=0\)
Solve \(x^4 - 13x^2 + 36 = 0\)
Spot it
Contains \(x^4\) and \(x^2\) — let \(u = x^2\)
Substitute
\(u^2 - 13u + 36 = 0\)
Factorise
\((u - 4)(u - 9) = 0\)
\(u = 4\) or \(u = 9\)
\(u = 4\) or \(u = 9\)
Back sub
\(x^2 = 4 \Rightarrow x = \pm 2\)
\(x^2 = 9 \Rightarrow x = \pm 3\)
\(x^2 = 9 \Rightarrow x = \pm 3\)
✓
\(x = \pm2\) or \(x = \pm3\)
Solve \(2\sqrt{x} - 5x^{1/2} + 3 = 0\) — wait! Rewrite first
Rewrite
\(2x - 7x^{1/2} + 3 = 0\) — actually this is \(2(\sqrt{x})^2 - 7\sqrt{x} + 3 = 0\)
Let
\(u = \sqrt{x}\) (so \(u \geq 0\))
Quadratic
\(2u^2 - 7u + 3 = 0\)
\((2u - 1)(u - 3) = 0\)
\(u = \tfrac{1}{2}\) or \(u = 3\)
\((2u - 1)(u - 3) = 0\)
\(u = \tfrac{1}{2}\) or \(u = 3\)
Back sub
\(\sqrt{x} = \tfrac{1}{2} \Rightarrow x = \tfrac{1}{4}\)
\(\sqrt{x} = 3 \Rightarrow x = 9\)
\(\sqrt{x} = 3 \Rightarrow x = 9\)
✓
\(x = \tfrac{1}{4}\) or \(x = 9\)
Handling "no real solution" cases
After substituting back, check whether each value of u is valid. Some values may be impossible:
If \(u = \sqrt{x}\), then \(u \geq 0\) — reject any negative u.
If \(u = e^x\), then \(u > 0\) — reject \(u \leq 0\).
If \(u = \sin x\), then \(-1 \leq u \leq 1\) — reject values outside this range.
Solve \(e^{2x} - 3e^x + 2 = 0\)
Let
\(u = e^x\), so \(u > 0\)
Quadratic
\(u^2 - 3u + 2 = (u-1)(u-2) = 0\)
\(u = 1\) or \(u = 2\)
\(u = 1\) or \(u = 2\)
Back sub
\(e^x = 1 \Rightarrow x = 0\)
\(e^x = 2 \Rightarrow x = \ln 2\)
\(e^x = 2 \Rightarrow x = \ln 2\)
✓
\(x = 0\) or \(x = \ln 2\)
Practice Questions
3 Easy · 3 Medium · 3 Hard — Equations, Inequalities & Modulus
Easy 1
[2] Solve \(|2x - 3| = 7\).
SOLUTION
Two cases
\(2x-3=7 \Rightarrow x=5\) OR \(2x-3=-7 \Rightarrow x=-2\)
\(\boldsymbol{x=5}\) or \(\boldsymbol{x=-2}\)
\(\boldsymbol{x=5}\) or \(\boldsymbol{x=-2}\)
Easy 2
[3] Solve the inequality \(x^2 - 2x - 8 \leq 0\).
SOLUTION
Factorise
\((x-4)(x+2)\leq0\). Roots: \(x=4,x=-2\).
Answer
U-shape below zero between roots: \(\boldsymbol{-2\leq x\leq4}\)
Easy 3
[2] Solve \(|3x + 1| = |x - 5|\).
SOLUTION
Case 1 (same sign)
\(3x+1=x-5 \Rightarrow 2x=-6 \Rightarrow x=-3\)
Case 2 (opp sign)
\(3x+1=-(x-5) \Rightarrow 4x=4 \Rightarrow x=1\)
Answer
\(\boldsymbol{x=-3}\) or \(\boldsymbol{x=1}\)
Medium 1
[4] Solve \(|x^2 - 4| = 3x\).
SOLUTION
Case 1
\(x^2-4=3x \Rightarrow x^2-3x-4=0 \Rightarrow (x-4)(x+1)=0 \Rightarrow x=4\) or \(x=-1\)
Check: \(x=4\): LHS=\(|12|=12\), RHS=12 ✓. \(x=-1\): RHS=\(-3<0\) ✗
Check: \(x=4\): LHS=\(|12|=12\), RHS=12 ✓. \(x=-1\): RHS=\(-3<0\) ✗
Case 2
\(x^2-4=-3x \Rightarrow x^2+3x-4=0 \Rightarrow (x+4)(x-1)=0 \Rightarrow x=-4\) or \(x=1\)
Check: \(x=1\): RHS=3, LHS=\(|-3|=3\) ✓. \(x=-4\): RHS=\(-12<0\) ✗
Check: \(x=1\): RHS=3, LHS=\(|-3|=3\) ✓. \(x=-4\): RHS=\(-12<0\) ✗
Answer
\(\boldsymbol{x=4}\) or \(\boldsymbol{x=1}\)
Medium 2
[4] Find the set of values of \(x\) satisfying \(\dfrac{x+1}{x-2} > 3\).
SOLUTION — never cross-multiply without knowing sign
Rearrange
\(\dfrac{x+1}{x-2}-3>0 \Rightarrow \dfrac{x+1-3(x-2)}{x-2}>0 \Rightarrow \dfrac{-2x+7}{x-2}>0\)
Critical values
Numerator=0: \(x=3.5\). Denominator=0: \(x=2\). Sign table: positive when \(\boldsymbol{2
Medium 3
[4] Solve \(|2x-1| \leq |x+3|\).
SOLUTION — square both sides (safe for ≤ with non-negatives)
Square
\((2x-1)^2\leq(x+3)^2\)
\(4x^2-4x+1\leq x^2+6x+9\)
\(3x^2-10x-8\leq0 \Rightarrow (3x+2)(x-4)\leq0\)
\(4x^2-4x+1\leq x^2+6x+9\)
\(3x^2-10x-8\leq0 \Rightarrow (3x+2)(x-4)\leq0\)
Answer
\(\boldsymbol{-\tfrac{2}{3}\leq x\leq4}\)
Hard 1
[5] Sketch \(y=|2x-6|\) and \(y=x+1\) on the same axes. Hence solve \(|2x-6|\leq x+1\).
SOLUTION — graphical approach
Intersections
Case 1: \(2x-6=x+1 \Rightarrow x=7\). Case 2: \(-(2x-6)=x+1 \Rightarrow x=\frac{5}{3}\)
Answer
Line is above/on V-shape between intersections: \(\boldsymbol{\tfrac{5}{3}\leq x\leq7}\)
Always sketch: draw V-shape of modulus, draw the line, shade where line is above/on the V.
Hard 2
[5] Find all values of \(k\) for which \(|4x-k|=2x+1\) has exactly one solution.
SOLUTION
Case 1 valid
\(4x-k=2x+1 \Rightarrow x=\frac{k+1}{2}\). Valid when \(4x\geq k\): \(2(k+1)\geq k \Rightarrow k\geq-2\)
Case 2 valid
\(-(4x-k)=2x+1 \Rightarrow x=\frac{k-1}{6}\). Valid when \(4x
One solution
Exactly one solution when only one case is valid: \(k<-2\) (only case 2) or \(k\geq2\) (only case 1).
\(\boldsymbol{k<-2}\) or \(\boldsymbol{k\geq2}\)
\(\boldsymbol{k<-2}\) or \(\boldsymbol{k\geq2}\)
Hard 3
[5] Given \(f(x)=2x^2-8x+5\), find the set of values of \(x\) for which \(f(x)\leq f(3)\).
SOLUTION
Find f(3)
\(f(3)=18-24+5=-1\)
Solve
\(2x^2-8x+5\leq-1 \Rightarrow 2x^2-8x+6\leq0 \Rightarrow x^2-4x+3\leq0 \Rightarrow (x-1)(x-3)\leq0\)
Answer
\(\boldsymbol{1\leq x\leq3}\)
Past Year Paper Questions
Cambridge 0606 — Equations, Inequalities & Graphs
0606
[10]
(a) Sketch \(y=|3x-6|\), stating the coordinates of the vertex. [2]
(b) Solve \(|3x-6|=x+2\). [4]
(c) Hence find the set of values of \(x\) for which \(|3x-6|>x+2\). [2]
(d) Find the value of \(k\) such that the equation \(|3x-6|=x+k\) has exactly one solution. [2]
(b) Solve \(|3x-6|=x+2\). [4]
(c) Hence find the set of values of \(x\) for which \(|3x-6|>x+2\). [2]
(d) Find the value of \(k\) such that the equation \(|3x-6|=x+k\) has exactly one solution. [2]
FULL SOLUTION
(a)
V-shape with vertex at \((2,0)\). y-intercept at \((0,6)\).
(b) Case 1
\(3x-6=x+2 \Rightarrow x=4\). Check: LHS=6, RHS=6 ✓
(b) Case 2
\(-(3x-6)=x+2 \Rightarrow -3x+6=x+2 \Rightarrow x=1\). Check: LHS=3, RHS=3 ✓
\(\boldsymbol{x=1}\) or \(\boldsymbol{x=4}\)
\(\boldsymbol{x=1}\) or \(\boldsymbol{x=4}\)
(c)
From sketch, V is above line outside the intersection points: \(\boldsymbol{x<1}\) or \(\boldsymbol{x>4}\)
(d)
One solution means line is tangent to V at vertex. At vertex \((2,0)\): \(0=2+k \Rightarrow \boldsymbol{k=-2}\)
0606 Style
Quadratic Discriminant
[9]
(a) Find the range of values of \(p\) for which \(x^2 + px + 2p - 3 = 0\) has two distinct real roots. [4]
(b) If \(p = 5\), solve the equation. [3]
(c) Find the value of \(p\) for which the equation has equal roots and state those roots. [2]
(b) If \(p = 5\), solve the equation. [3]
(c) Find the value of \(p\) for which the equation has equal roots and state those roots. [2]
FULL WORKED SOLUTION
(a)
\(\Delta=p^2-4(2p-3)>0\)
\(p^2-8p+12>0\Rightarrow(p-2)(p-6)>0\)
\(\boldsymbol{p<2}\) or \(\boldsymbol{p>6}\)
\(p^2-8p+12>0\Rightarrow(p-2)(p-6)>0\)
\(\boldsymbol{p<2}\) or \(\boldsymbol{p>6}\)
(b) p=5
\(x^2+5x+7=0\). \(\Delta=25-28=-3<0\) — no real roots. (Check \(p=5\) is between 2 and 6.).
(c)
Equal roots: \(p=2\) or \(p=6\).
If \(p=2\): \(x^2+2x+1=0\Rightarrow(x+1)^2=0\Rightarrow\boldsymbol{x=-1}\)
If \(p=6\): \(x^2+6x+9=0\Rightarrow(x+3)^2=0\Rightarrow\boldsymbol{x=-3}\)
If \(p=2\): \(x^2+2x+1=0\Rightarrow(x+1)^2=0\Rightarrow\boldsymbol{x=-1}\)
If \(p=6\): \(x^2+6x+9=0\Rightarrow(x+3)^2=0\Rightarrow\boldsymbol{x=-3}\)
0606 Style
System of Non-linear Equations
[6]
Solve the following simultaneous equations:
\(y = 2x + 1\) and \(y = x^2 - 2x + 5\)
Give your answers in exact form and state what the results tell you geometrically. [6]
\(y = 2x + 1\) and \(y = x^2 - 2x + 5\)
Give your answers in exact form and state what the results tell you geometrically. [6]
FULL WORKED SOLUTION
Equate
\(2x+1=x^2-2x+5\Rightarrow x^2-4x+4=0\Rightarrow(x-2)^2=0\)
Solve
\(x=2\) (repeated). \(y=2(2)+1=5\). Point: \(\boldsymbol{(2,5)}\)
Geometric
One repeated intersection point means the line \(y=2x+1\) is a tangent to the parabola at \((2,5)\).