Equations, inequalities and graphs

1. Overview

Topic 4, Equations, Inequalities, and Graphs, is a cornerstone of the Additional Mathematics syllabus. It extends your IGCSE Maths skills by introducing the modulus function, hidden quadratics, and cubic polynomials. Expect to see these concepts tested in both Paper 1 (without a calculator, emphasising algebraic manipulation) and Paper 2. Mastery of this topic is crucial for success in coordinate geometry, calculus, and trigonometric functions. You'll need to be adept at solving modulus equations and inequalities, using substitution to solve hidden quadratics, and sketching cubic graphs (including their moduli). Pay close attention to exact values, sign errors, and domain restrictions, as these are common pitfalls highlighted in examiner reports.

Key Definitions

• Modulus (Absolute Value): Written as $|x|$, it represents the magnitude of a number regardless of its sign. $|x| = x$ if $x \geq 0$ and $|x| = -x$ if $x < 0$.
• Hidden Quadratic: An equation that does not initially look like a quadratic but can be transformed into $au^2 + bu + c = 0$ using a clever substitution (e.g., $u = e^x$ or $u = x^{1/n}$).
• Critical Values: The specific values of $x$ where an inequality changes from being true to false (usually the roots of the corresponding equation).
• Extraneous Solution: A solution obtained through valid algebraic steps (like squaring) that does not actually satisfy the original equation when substituted back.

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Core Content

3.1 Linear Modulus Equations

To solve $|f(x)| = g(x)$, you must consider two cases: $f(x) = g(x)$ and $f(x) = -g(x)$. Crucial: When $g(x)$ contains a variable (e.g., $|ax+b| = cx+d$), you must check your answers to ensure they don't result in a negative value on the non-modulus side.

Worked example 1 — Solving $|3x - 1| = 2x + 5$

Question: Solve the equation $|3x - 1| = 2x + 5$.

1. Case 1: $3x - 1 = 2x + 5$ Justification: Assume the expression inside the modulus is positive.
2. $3x - 2x = 5 + 1$ Justification: Rearrange the equation to isolate $x$.
3. $x = 6$ Justification: Simplify.
4. Case 2: $3x - 1 = -(2x + 5)$ Justification: Assume the expression inside the modulus is negative.
5. $3x - 1 = -2x - 5$ Justification: Expand the bracket.
6. $3x + 2x = -5 + 1$ Justification: Rearrange the equation to isolate $x$.
7. $5x = -4$ Justification: Simplify.
8. $x = -\frac{4}{5}$ Justification: Divide both sides by 5.
9. Check: For $x=6$: $|3(6)-1| = |18-1| = |17| = 17$ and $2(6)+5 = 12+5 = 17$ (Valid) Justification: Substitute $x=6$ into the original equation to verify.
10. For $x=-\frac{4}{5}$: $|3(-\frac{4}{5})-1| = |-\frac{12}{5}-\frac{5}{5}| = |-\frac{17}{5}| = \frac{17}{5}$; $2(-\frac{4}{5})+5 = -\frac{8}{5} + \frac{25}{5} = \frac{17}{5}$ (Valid) Justification: Substitute $x=-\frac{4}{5}$ into the original equation to verify.

Final Answer: $x = 6, x = -\frac{4}{5}$

Worked example 2 — Solving $|x^2 - 5| = 4$

Question: Solve the equation $|x^2 - 5| = 4$.

1. Case 1: $x^2 - 5 = 4$ Justification: Assume the expression inside the modulus is positive.
2. $x^2 = 9$ Justification: Add 5 to both sides.
3. $x = \pm 3$ Justification: Take the square root of both sides.
4. Case 2: $x^2 - 5 = -4$ Justification: Assume the expression inside the modulus is negative.
5. $x^2 = 1$ Justification: Add 5 to both sides.
6. $x = \pm 1$ Justification: Take the square root of both sides.

Final Answer: $x = \pm 3, x = \pm 1$

Worked example 3 — Solving $|2x-1| = |x+3|$

Question: Solve the equation $|2x-1| = |x+3|$.

1. Case 1: $2x - 1 = x + 3$ Justification: Assume both expressions inside the modulus are positive.
2. $2x - x = 3 + 1$ Justification: Rearrange the equation to isolate $x$.
3. $x = 4$ Justification: Simplify.
4. Case 2: $2x - 1 = -(x + 3)$ Justification: Assume the first expression is positive and the second is negative.
5. $2x - 1 = -x - 3$ Justification: Expand the bracket.
6. $2x + x = -3 + 1$ Justification: Rearrange the equation to isolate $x$.
7. $3x = -2$ Justification: Simplify.
8. $x = -\frac{2}{3}$ Justification: Divide both sides by 3.
9. Case 3: $-(2x - 1) = x + 3$ Justification: Assume the first expression is negative and the second is positive.
10. $-2x + 1 = x + 3$ Justification: Expand the bracket.
11. $-2x - x = 3 - 1$ Justification: Rearrange the equation to isolate $x$.
12. $-3x = 2$ Justification: Simplify.
13. $x = -\frac{2}{3}$ Justification: Divide both sides by -3.
14. Case 4: $-(2x - 1) = -(x + 3)$ Justification: Assume both expressions inside the modulus are negative.
15. $-2x + 1 = -x - 3$ Justification: Expand the bracket.
16. $-2x + x = -3 - 1$ Justification: Rearrange the equation to isolate $x$.
17. $-x = -4$ Justification: Simplify.
18. $x = 4$ Justification: Multiply both sides by -1.

Final Answer: $x = 4, x = -\frac{2}{3}$

3.2 Modulus Inequalities

There are two main methods:

1. Algebraic (Squaring): Useful for $|f(x)| \leq |g(x)|$ because $|a|^2 = a^2$.
2. Graphical/Critical Values: Find where the equations are equal, then test values in each region.

Worked example — Solve $|2x + 1| < |x - 4|$

Question: Find the set of values of $x$ for which $|2x + 1| < |x - 4|$.

1. Square both sides: $(2x + 1)^2 < (x - 4)^2$ Justification: Squaring eliminates the modulus signs.
2. Expand: $4x^2 + 4x + 1 < x^2 - 8x + 16$ Justification: Expand the brackets.
3. Rearrange to a quadratic inequality: $3x^2 + 12x - 15 < 0$ Justification: Move all terms to one side.
4. Divide by 3: $x^2 + 4x - 5 < 0$ Justification: Simplify the inequality.
5. Factorise: $(x+5)(x-1) < 0$ Justification: Factorise the quadratic expression.
6. Find roots: $x = -5, x = 1$. Since it is "$< 0$", we take the region between the roots. Justification: The inequality is satisfied when one factor is positive and the other is negative.

Final Answer: $-5 < x < 1$

3.3 Hidden Quadratics (Substitution)

Look for a middle term where the power is exactly half of the leading term's power.

Worked example — Solve $3e^x = 12 - 5e^{-x}$

Question: Solve the equation $3e^x = 12 - 5e^{-x}$. Give your answer in exact form.

1. Multiply everything by $e^x$: $3e^{2x} = 12e^x - 5$ Justification: Multiply by $e^x$ to eliminate the negative exponent.
2. Rearrange: $3(e^x)^2 - 12e^x + 5 = 0$ Justification: Rearrange into a quadratic form.
3. Let $u = e^x$: $3u^2 - 12u + 5 = 0$ Justification: Substitute $u$ for $e^x$ to form a quadratic equation.
4. Use Quadratic Formula: $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)} = \frac{12 \pm \sqrt{144 - 60}}{6} = \frac{12 \pm \sqrt{84}}{6}$ Justification: Apply the quadratic formula to solve for $u$.
5. Simplify surd: $\sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21} \implies u = \frac{12 \pm 2\sqrt{21}}{6} = 2 \pm \frac{\sqrt{21}}{3}$ Justification: Simplify the square root.
6. Solve for $x$: $e^x = 2 \pm \frac{\sqrt{21}}{3} = \frac{6 \pm \sqrt{21}}{3} \implies x = \ln\left(\frac{6 \pm \sqrt{21}}{3}\right)$ Justification: Substitute back $e^x$ for $u$ and take the natural logarithm of both sides.

Final Answer: $x = \ln\left(\frac{6 + \sqrt{21}}{3}\right), x = \ln\left(\frac{6 - \sqrt{21}}{3}\right)$ (Exact values required).

Worked example — Solve $x^{4/3} + x^{2/3} - 12 = 0$

Question: Solve the equation $x^{4/3} + x^{2/3} - 12 = 0$.

1. Let $u = x^{2/3}$: $u^2 + u - 12 = 0$ Justification: Substitute $u$ for $x^{2/3}$ to form a quadratic equation.
2. Factorise: $(u+4)(u-3) = 0$ Justification: Factorise the quadratic expression.
3. Find roots: $u = -4, u = 3$ Justification: Solve for $u$.
4. Solve for $x$: $x^{2/3} = -4$ or $x^{2/3} = 3$ Justification: Substitute back $x^{2/3}$ for $u$.
5. Consider $x^{2/3} = -4$: $(x^{1/3})^2 = -4$. Since a real number squared cannot be negative, there are no real solutions in this case. Justification: Recognise that a real number raised to an even power cannot be negative.
6. Consider $x^{2/3} = 3$: $(x^{1/3})^2 = 3 \implies x^{1/3} = \pm \sqrt{3}$ Justification: Take the square root of both sides.
7. Cube both sides: $x = (\pm \sqrt{3})^3 = \pm 3\sqrt{3}$ Justification: Cube both sides to solve for $x$.

Final Answer: $x = 3\sqrt{3}, x = -3\sqrt{3}$

3.4 Cubic Graphs and their Moduli

When sketching $y = |f(x)|$ where $f(x)$ is a cubic:

1. Sketch $y = f(x)$ first.
2. Label x-intercepts (roots) and the y-intercept.
3. Reflect any part of the curve below the x-axis upwards to make it positive.
4. The "sharp turns" on the x-axis are where the modulus acts.

3.5 Solving Cubic Inequalities Graphically

To solve $f(x) \geq d$:

1. Sketch $y = f(x)$.
2. Draw the horizontal line $y = d$.
3. Identify the x-intervals where the cubic curve is on or above the line $y = d$.

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Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations & Formulas

Equation Type	General Form	Notes
Modulus Definition	$	x
Hidden Quadratic	$ax^{2n} + bx^n + c = 0$	Let $u = x^n$.
Logarithmic Substitution	$a(\ln x)^2 + b\ln x + c = 0$	Let $u = \ln x$; ensure $x > 0$.
Modulus Inequality	$	f(x)

Modulus Definition: $|x| = \begin{cases} x, & \text{if } x \geq 0 \ -x, & \text{if } x < 0 \end{cases}$

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Common Mistakes to Avoid

• ❌ Wrong: Solving $|2x - 5| = x - 10$ and accepting all answers without checking for extraneous solutions.
• ✓ Right: Always substitute your answers back into the original equation. If the result is $|positive| = negative$, that solution is extraneous and must be rejected.
• ❌ Wrong: Leaving answers as decimals like $x = 0.447$ when the question requires exact values.
• ✓ Right: Use exact values: $x = \frac{\sqrt{5}}{5}$ or $x = \ln 2$. If the question involves surds or logarithms, express your answer in terms of surds or logarithms unless otherwise specified.
• ❌ Wrong: Forgetting to reflect the $y$-intercept when sketching $y = |f(x)|$ if the original intercept was negative.
• ✓ Right: If $f(0) = -4$, then for the modulus graph, the $y$-intercept must be $(0, 4)$. Remember to reflect the portion of the graph below the x-axis above the x-axis.
• ❌ Wrong: Solving $x^{2/3} + x^{1/3} - 6 = 0$ and saying $x = 2$ or $x = -3$.
• ✓ Right: Those are values for $u$. You must solve $x^{1/3} = 2 \implies x = 8$ and $x^{1/3} = -3 \implies x = -27$. Always substitute back to find the value of the original variable.
• ❌ Wrong: Forgetting to consider the domain of logarithmic functions when solving equations involving $\ln x$.
• ✓ Right: Remember that $\ln x$ is only defined for $x > 0$. Any solution where $x \leq 0$ must be rejected immediately.
• ❌ Wrong: Making sign errors when expanding brackets or factorising quadratic expressions.
• ✓ Right: Circle every sign and bracket as you check your work. Pay close attention to the signs when expanding and factorising.
• ❌ Wrong: Not showing sufficient working in "show that" questions, especially when factorising quadratics or using the quadratic formula.
• ✓ Right: Demonstrate every intermediate step, no matter how obvious it seems to you. Show the factorisation or the substitution into the quadratic formula.

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Exam Tips

• Show Every Step: In "Show that" questions, don't skip from a quadratic to its roots. Show the factorisation or the substitution into the quadratic formula.
• Check Domain Restrictions: For equations involving $\ln x$, any solution where $x \leq 0$ must be rejected immediately as $\ln$ is undefined for non-positive numbers.
• Sketching Labels: Examiners look for four specific things on a cubic sketch: Correct shape, $x$-intercepts, $y$-intercept, and clear reflection of the modulus parts.
• Command Words:
  • "Solve": Find the values of $x$.
  • "Find the set of values": Your answer must be an inequality (e.g., $x < 2, x > 5$).
• Paper 1 Strategy: Since it is non-calculator, if you end up with a very complex square root that doesn't simplify, re-check your expansion of brackets or sign signs earlier in the problem.
• Paper 2 Strategy: Use your calculator to check your algebraic solutions, especially for hidden quadratics and modulus equations. Be mindful of storing intermediate values to maintain accuracy.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

(a) Solve the equation $|2x - 1| = |x + 4|$. [4]

(b) Find the value of $k$ for which the equation $x^2 + (k - 2)x + 4 = 0$ has equal roots. [4]

Worked Solution:

(a)

1. Consider the two cases where the expressions inside the absolute value signs are equal or negatives of each other. $2x - 1 = x + 4$ or $2x - 1 = -(x + 4)$ [Break the absolute value into two linear equations]

2. Solve the first equation. $2x - x = 4 + 1 \implies x = 5$ [Isolate x]

3. Solve the second equation. $2x - 1 = -x - 4 \implies 3x = -3 \implies x = -1$ [Isolate x]

4. State both solutions. $\boxed{x = 5 \text{ or } x = -1}$ [Final answers]

How to earn full marks: Remember to consider both positive and negative cases when dealing with absolute values, and clearly state both solutions.

(b)

1. For equal roots, the discriminant must be equal to zero. $(k-2)^2 - 4(1)(4) = 0$ [Apply the discriminant formula $b^2-4ac = 0$]

2. Expand and simplify the equation. $k^2 - 4k + 4 - 16 = 0 \implies k^2 - 4k - 12 = 0$ [Expand the squared term and simplify]

3. Factorize the quadratic equation. $(k - 6)(k + 2) = 0$ [Factorize the equation]

4. Solve for $k$. $k = 6$ or $k = -2$ [State both solutions]

How to earn full marks: Set the discriminant to zero, show your working in expanding and factorizing, and state both values of k.

Common Pitfall: Remember that absolute value equations often have two solutions. Don't forget to consider both the positive and negative cases. Also, when finding the values of $k$ for equal roots, make sure you set the discriminant equal to zero and solve the resulting quadratic equation carefully.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

(a) Solve the inequality $|3x - 2| \leq 4$. [4]

(b) Given that $x = 2 + \sqrt{3}$, find the value of $x + \frac{1}{x}$ in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers. [3]

Worked Solution:

(a)

1. Express the inequality without the absolute value. $-4 \leq 3x - 2 \leq 4$ [Transform the absolute value inequality to a compound inequality]

2. Add 2 to all parts of the inequality. $-2 \leq 3x \leq 6$ [Isolate the term with x]

3. Divide all parts of the inequality by 3. $-\frac{2}{3} \leq x \leq 2$ [Isolate x]

4. State the solution. $\boxed{-\frac{2}{3} \leq x \leq 2}$ [Final answer]

How to earn full marks: Transform the absolute value inequality into a compound inequality and isolate x in all parts to find the solution.

(b)

1. Find the reciprocal of $x$. $\frac{1}{x} = \frac{1}{2 + \sqrt{3}}$ [State the reciprocal]

2. Rationalize the denominator. $\frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$ [Multiply by the conjugate to rationalize]

3. Calculate $x + \frac{1}{x}$. $x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$ [Add x and its reciprocal]

4. State the solution $\boxed{4}$ [Final answer]

How to earn full marks: Remember to rationalize the denominator by multiplying by the conjugate, and simplify to get the final answer.

Common Pitfall: When solving absolute value inequalities, remember to split the problem into two separate inequalities. Also, when rationalizing the denominator, be careful with your signs and remember to multiply both the numerator and denominator by the conjugate.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [11 marks]

Question:

(a) Solve the equation $3e^{2x} - 10e^x + 3 = 0$. Give your answers correct to 3 significant figures. [5]

(b) (i) Sketch the graph of $y = (x+1)(x-2)(x-4)$, showing the points where the graph intersects the $x$-axis and the $y$-axis. [3]

(ii) Hence, solve the inequality $(x+1)(x-2)(x-4) < 0$. [3]

Worked Solution:

(a)

1. Substitute $y = e^x$ into the equation. $3y^2 - 10y + 3 = 0$ [Substitution to form quadratic equation]

2. Factorize the quadratic equation. $(3y - 1)(y - 3) = 0$ [Factorize the equation]

3. Solve for $y$. $y = \frac{1}{3}$ or $y = 3$ [Solve for y]

4. Substitute back $e^x = y$ and solve for $x$. $e^x = \frac{1}{3} \implies x = \ln\left(\frac{1}{3}\right) = -\ln 3$ or $e^x = 3 \implies x = \ln 3$ [Solve for x]

5. Write the solutions to 3 sf. $\boxed{x = -1.10 \text{ or } x = 1.10}$ [Final answers]

How to earn full marks: Use a substitution to form a quadratic, solve for the substituted variable, then substitute back to find x, giving answers to 3 s.f.

(b) (i)

1. Identify the $x$-intercepts. The graph intersects the $x$-axis at $x = -1, 2, 4$. [State the roots of the cubic]

2. Identify the $y$-intercept. When $x = 0$, $y = (1)(-2)(-4) = 8$. The graph intersects the $y$-axis at $y = 8$. [Find the y-intercept]

3. Sketch the graph showing intercepts.

   📊A cubic curve with x-intercepts at -1, 2, and 4, and a y-intercept at 8. The curve starts from negative infinity in the bottom left, crosses the x-axis at x=-1, reaches a local maximum between -1 and 2, crosses the x-axis at x=2, reaches a local minimum between 2 and 4, crosses the x-axis at x=4, and continues to positive infinity in the top right. The axes are labeled x and y.

How to earn full marks: Clearly label all x and y intercepts on your sketch, and ensure the shape of the cubic is correct.

(ii)

1. Identify the intervals where the graph is below the $x$-axis. From the sketch, $(x+1)(x-2)(x-4) < 0$ when $x < -1$ or $2 < x < 4$. [Interpret the graph]

2. State the solution to the inequality. $\boxed{x < -1 \text{ or } 2 < x < 4}$ [Final answer]

How to earn full marks: Use your sketch to identify the correct intervals where the graph is below the x-axis, and write the solution as inequalities.

Common Pitfall: When solving equations involving $e^x$, remember to use a substitution to simplify the equation into a quadratic form. Also, when sketching cubic graphs, make sure to clearly label all intercepts and consider the overall shape of the curve. When solving inequalities based on the graph, pay close attention to whether the inequality is strict ($<$) or non-strict ($\leq$).

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [10 marks]

Question:

(a) Find the range of values of $x$ for which $|2x - 3| > x + 1$. [6]

(b) The function $f$ is defined by $f(x) = x^{4/3} - 5x^{2/3} + 4$ for $x \geq 0$.

(i) By using a suitable substitution, solve the equation $f(x) = 0$. [4]

Worked Solution:

(a)

1. Express the inequality without the absolute value. $2x - 3 > x + 1$ or $2x - 3 < -(x + 1)$ [Transform the absolute value inequality into two separate inequalities]

2. Solve the first inequality. $2x - x > 1 + 3 \implies x > 4$ [Solve for x]

3. Solve the second inequality. $2x - 3 < -x - 1 \implies 3x < 2 \implies x < \frac{2}{3}$ [Solve for x]

4. Check for extraneous solutions. Since $x+1$ must be non-negative for the absolute value inequality to hold, $x>-1$

5. Combine the solutions and exclude any extraneous solutions. $\boxed{x > 4 \text{ or } -1 < x < \frac{2}{3}}$ [Final answer]

How to earn full marks: Remember to consider both cases of the absolute value, solve each inequality, and check for extraneous solutions by considering the domain.

(b) (i)

1. Substitute $y = x^{2/3}$ into the equation. $y^2 - 5y + 4 = 0$ [Substitution to form quadratic equation]

2. Factorize the quadratic equation. $(y - 4)(y - 1) = 0$ [Factorize the equation]

3. Solve for $y$. $y = 4$ or $y = 1$ [Solve for y]

4. Substitute back $x^{2/3} = y$ and solve for $x$. $x^{2/3} = 4 \implies x = 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$ or $x^{2/3} = 1 \implies x = 1^{3/2} = 1$ [Solve for x]

5. State the solutions. $\boxed{x = 8 \text{ or } x = 1}$ [Final answer]

How to earn full marks: Use the correct substitution to form a quadratic equation, solve for the substituted variable, and then substitute back to find the values of x.

Common Pitfall: When solving absolute value inequalities, it's crucial to check for extraneous solutions, especially when the expression on the other side of the inequality involves a variable. Also, when dealing with equations involving fractional exponents, remember to use a suitable substitution to transform the equation into a more manageable form, like a quadratic.