Equations, inequalities and graphs

A modulus is never negative. The equation |ax + b| = c with c geqslant 0 means that ax + b is c or -c.

Form the pair of linear equations, write both down, then solve each one.

For |ax + b| = cx + d, form the same pair. Then check each answer in the original equation. The right-hand side cannot be negative, so an answer that makes it negative is rejected.

When both sides have a modulus, the two expressions are equal or they are opposites. Write the pair of linear equations and solve each.

The other method is to square both sides. That gives a quadratic with no modulus. Square the whole of each side. Every term and every number is squared. Then solve the quadratic and give both of its solutions.

The graph of y = |f(x)| has no part below the x-axis. Any part of y = f(x) that is below the axis is reflected above it.

For y = |ax + b| the graph is a V shape. Its vertex is on the x-axis. Use a ruler for the straight sections and mark the intercepts with both axes.

To solve graphically, draw each graph accurately. The x-values where the graphs cross are the solutions of an equation. For an inequality, read the x-values where one graph is above or below the other.

Solve the matching equation first. Its answers are the critical values. Then decide which regions satisfy the inequality.

If |f(x)| > c, the values lie outside the critical values: two separate regions. If |f(x)| < c, they lie between: one region. Write each region as its own inequality.

A sketch makes the regions clear. When both sides are a modulus, as in |ax + b| leqslant |cx + d|, you can square both sides and solve the quadratic inequality. Give the final answer as inequalities in x.

|ax² + bx + c| = d means ax² + bx + c = d or ax² + bx + c = -d. Solve both quadratics. There can be up to four solutions.

For |ax² + bx + c| > d, solve ax² + bx + c > d or ax² + bx + c < -d. For |ax² + bx + c| leqslant d, both ax² + bx + c leqslant d and ax² + bx + c geqslant -d must hold.

The graph of y = |ax² + bx + c| is a smooth curve. Where the quadratic dips below the x-axis, that part is reflected, so the graph has a point (a cusp) on the axis at each root. Mark the intercepts with both axes.

Some equations are a disguised quadratic. Spot the repeated expression, such as 2^(x), x^(2/3), ln 5x or e^(x). Let it be y and rewrite the equation.

Solve the quadratic in y. Then go back to the original variable: put each value of y back into the substitution. Reject a value that gives no solution, such as 2^(x) = -20.

The syllabus examples show the substitution to look for. You are expected to identify the appropriate substitution yourself. Taking logarithms of every term does not work, so look for the repeated expression first.

For y = (x - a)(x - b)(x - c), the x-intercepts are a, b and c. Put x = 0 for the y-intercept.

Draw one smooth curve through the intercepts. A cubic with a positive x³ term rises on the right and falls on the left. Extend the arms beyond the outer roots and do not bend them in.

For y = |f(x)|, reflect the parts below the x-axis. Each root where the curve crossed the axis becomes a cusp. Label every intercept.

To solve f(x) geqslant d, sketch y = f(x) and draw the line y = d. The line meets the curve at the critical values. On a graph you are given, read them off the graph: do not use a calculator, and do not give more figures than the grid allows.

For > or geqslant, take the x-values where the curve is above the line. For < or leqslant, take those where it is below. A cubic often gives two regions, so write each as an inequality.

Use < and > when the question does, and leqslant and geqslant when it does. Write a single region between two values as one inequality such as 2 < x < 3.