Equations, inequalities and graphs

📚The mathematics

Modulus equations

\( |X| = c \) (with \( c \ge 0 \)) means \( X = c \) or \( X = -c \). Every modulus equation is two ordinary equations; solve both.

• \( |ax + b| = cx + d \): solve \( ax + b = cx + d \) and \( ax + b = -(cx + d) \), then check each answer in the original. The right-hand side can be negative, and a modulus never is, so one of the two answers may be false.
• \( |ax + b| = |cx + d| \): both sides are non-negative, so squaring is safe: \( (ax + b)^{2} = (cx + d)^{2} \) gives a quadratic. (Or solve \( ax + b = \pm(cx + d) \).)
• \( |ax^{2} + bx + c| = d \): two quadratics, \( = d \) and \( = -d \). Either can have 0, 1 or 2 roots, so there can be up to four solutions. A sketch tells you how many to expect.

Modulus inequalities

Graphically: sketch both sides, find where they meet (by algebra, not by reading the graph), and read off where one is above the other. Algebraically, two facts do most of the work:

Equations that are quadratics in disguise

If one term’s power is twice another’s, substitute \(u\) for the smaller one:

The syllabus says you are expected to find the substitution yourself. After solving for \(u\), undo it — and reject values it cannot take (\( e^{x} \) and \( 2^{x} \) are always positive).

Cubic graphs and their moduli

For \( y = k(x - a)(x - b)(x - c) \): it crosses the \(x\)-axis at \(a\), \(b\) and \(c\), the \(y\)-intercept is found by putting \( x = 0 \), and the ends go bottom-left to top-right if \( k > 0 \) (top-left to bottom-right if \( k < 0 \)). For a factor with a coefficient, such as \( 2x + 5 \), the root is \( -\tfrac{5}{2} \), and the leading coefficient is the product of the \(x\)-coefficients.

For \( y = |f(x)| \), reflect the parts below the axis upwards, exactly as in Topic 1. Label all the intercepts: that is what the marks are for.

Cubic inequalities

To solve \( f(x) \ge d \): find where \( f(x) = d \), draw \( y = d \) on the sketch, and read off where the curve is on or above it. Because a cubic can cross a horizontal line three times, the answer is usually two pieces — a bounded interval and a ray. In the figure, \( (x + 2)(x - 1)(x - 3) = 6 \) becomes \( x^{3} - 2x^{2} - 5x = 0 \), so \( x = 0 \) or \( x = 1 \pm \sqrt{6} \).

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) Solve \( |3x - 2| = 7 \). [2]

\( 3x - 2 = 7 \Rightarrow x = 3 \); \( 3x - 2 = -7 \Rightarrow x = -\tfrac{5}{3} \).

2. (No calculator.) Solve \( |x + 4| = |3x - 2| \). [3]

Square: \( x^{2} + 8x + 16 = 9x^{2} - 12x + 4 \Rightarrow 8x^{2} - 20x - 12 = 0 \Rightarrow 2x^{2} - 5x - 3 = 0 \Rightarrow (2x + 1)(x - 3) = 0 \). So \( x = 3 \) or \( x = -\tfrac{1}{2} \). (Check: \( |7| = |7| \) and \( \left|\tfrac{7}{2}\right| = \left|-\tfrac{7}{2}\right| \).)

3. (No calculator.) Solve \( |x^{2} - 5x| = 6 \). [4]

\( x^{2} - 5x - 6 = 0 \Rightarrow (x - 6)(x + 1) = 0 \Rightarrow x = 6, -1 \). \( x^{2} - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0 \Rightarrow x = 2, 3 \). Four solutions: \( -1, 2, 3, 6 \). The second case works because the reflected hump of \( |x^{2} - 5x| \) reaches \( 6.25 \), above 6.

4. (No calculator.) Solve \( 3|2x - 1| > 9 \). [3]

\( |2x - 1| > 3 \), so \( 2x - 1 > 3 \) or \( 2x - 1 < -3 \): \( x > 2 \) or \( x < -1 \).

5. (No calculator.) Using a suitable substitution, solve \( x^{\frac{2}{3}} - 5x^{\frac{1}{3}} + 4 = 0 \). [4]

Let \( u = x^{\frac{1}{3}} \): \( u^{2} - 5u + 4 = 0 \Rightarrow (u - 1)(u - 4) = 0 \). So \( x^{\frac{1}{3}} = 1 \Rightarrow x = 1 \), or \( x^{\frac{1}{3}} = 4 \Rightarrow x = 64 \).

6. (Calculator.) Solve \( 2e^{x} + 3e^{-x} = 7 \). [4]

Let \( u = e^{x} \) and multiply by \(u\): \( 2u^{2} - 7u + 3 = 0 \Rightarrow (2u - 1)(u - 3) = 0 \). So \( e^{x} = 3 \Rightarrow x = \ln 3 = 1.10 \), or \( e^{x} = \tfrac{1}{2} \Rightarrow x = -\ln 2 = -0.693 \).

7. (No calculator.) The curve \( y = (x + 1)(x - 2)(x - 4) \) has turning points at approximately \( (0.21, 8.21) \) and \( (3.12, -4.06) \). Sketch \( y = |(x + 1)(x - 2)(x - 4)| \), stating the intercepts with the axes, and hence state the number of solutions of \( |(x + 1)(x - 2)(x - 4)| = 5 \). [4]

\(x\)-intercepts \( (-1, 0) \), \( (2, 0) \), \( (4, 0) \); \(y\)-intercept \( (0, 8) \). The part for \( x < -1 \) and the part between 2 and 4 are reflected, so the second hump reaches only about 4.06. The line \( y = 5 \) meets the left branch once, the hump between \(-1\) and 2 twice (it rises to 8.21), misses the small hump, and meets the right branch once: 4 solutions.