1The idea in one paragraph

To solve ax² + bx + c = 0 is to find where the parabola y = ax² + bx + c meets the x-axis. There are three methods. Factorising is fastest when it works. Completing the square always works and gives exact answers. The quadratic formula is completing the square done once and for all, so it always works too. Before you solve anything, the discriminant, Δ = b² − 4ac, tells you how many real roots there are: two if Δ > 0, one repeated root if Δ = 0, none if Δ < 0. That makes it the tool for questions with an unknown coefficient k. A quadratic inequality is solved by finding the roots first and then reading from a sketch where the parabola is above or below the axis.

2Roots, zeros and solutions

The guide uses three words for the same numbers. The roots or solutions of the equation ax² + bx + c = 0 are the values of x that make it true. The zeros of the function f(x) = ax² + bx + c are the inputs that give output zero. On the graph, both are the x-coordinates of the x-intercepts. A quadratic has at most two of them, because a parabola can cross a horizontal line at most twice.

Every method below starts from the same place: get zero on one side. The equation x² = 3x + 10 must become x² − 3x − 10 = 0 before you do anything else.

3Solving by factorising

If a product of two numbers is zero, one of them must be zero. That fact, the zero product rule, turns a factorised quadratic into two linear equations.

When a ≠ 1, look for the pair of numbers with product ac and sum b, and split the middle term.

A trap to avoid: from x² = 3x, dividing both sides by x gives x = 3 and loses a root. Rearrange instead: x² − 3x = 0, so x(x − 3) = 0, and x = 0 or x = 3. Never divide by an expression that could be zero.

Factorising only works when the roots are rational, which is less often than exam practice makes it seem. If you cannot find the factors in half a minute, switch method.

4Solving by completing the square

The vertex form of 2.6 turns any quadratic equation into "something squared equals a number", which you undo with a square root, remembering both signs.

These are the exact roots. As decimals they are 0.464 and −6.46, but a Paper 1 question that says "exact" wants −3 ± 2√3, and a decimal loses the accuracy mark.

Figure 1 shows why the method has its name. The expression x² + 6x is a square of side x plus two strips of 3 by x. Add the missing 3-by-3 corner, 9, and you have a complete square of side x + 3.

5The quadratic formula

Completing the square on the general equation ax² + bx + c = 0 gives a formula for the roots of every quadratic. It is in the formula booklet:

x = (−b ± √(b² − 4ac)) / 2a, for ax² + bx + c = 0, a ≠ 0.

Use it by writing a, b and c down first, with their signs, and then substituting.

The two sign errors that cost the most marks are both in that second line: −b when b is already negative (−(−3) = +3), and −4ac when c is negative (−4 × 2 × (−4) = +32). Brackets around every substituted value prevent both.

Where the formula comes from. The guide lists this as enrichment, and seeing it once makes the formula easier to trust. Divide by a, complete the square, and solve:

The formula also shows why the axis of symmetry is x = −b/2a (2.6): the two roots are −b/2a plus and minus the same amount, so they sit symmetrically about it. People have solved quadratic problems for a very long time; the guide mentions methods in Babylonian mathematics and in the ancient Indian Sulba Sutras and Bakhshali manuscript.

Which method? Look before you choose.

6The discriminant and the nature of the roots

The part of the formula under the square root, b² − 4ac, is called the discriminant and written Δ. It decides everything about the roots, because you can only take the square root of a number that is not negative.

• Δ > 0: two distinct real roots. The ± gives two different values. The parabola crosses the x-axis twice.
• Δ = 0: two equal real roots (one repeated root, x = −b/2a). The ± adds and subtracts zero. The parabola touches the x-axis at its vertex.
• Δ < 0: no real roots. There is no real square root of a negative number. The parabola misses the x-axis entirely.

Figure 2 shows all three for parabolas that differ only in c. For y = x² − 4x + 1, Δ = 16 − 4 = 12. For y = x² − 4x + 4, Δ = 16 − 16 = 0. For y = x² − 4x + 7, Δ = 16 − 28 = −12.

Two more facts are useful. If Δ is a perfect square (0, 1, 4, 9, 25, 49 …) and a, b, c are whole numbers, the roots are rational and the quadratic factorises; for 6x² + x − 2, Δ = 1 + 48 = 49 = 7², which is why section 3 found factors. And if a > 0 and Δ < 0, the parabola sits entirely above the x-axis, so the quadratic is positive for every x. That is how you show, for example, that x² − 2x + 5 > 0 for all x: a = 1 > 0 and Δ = 4 − 20 = −16 < 0.

The phrase "real roots" matters. When Δ < 0 there are roots that are complex numbers, which HL students meet in 1.12. At standard level, "no real roots" is the whole answer.

7Quadratic inequalities

A quadratic inequality such as x² − x − 12 > 0 asks for every x where the parabola is above the x-axis. The method is always the same three steps:

1. Rearrange so that one side is zero.
2. Solve the equation to find the critical values, the roots.
3. Sketch the parabola through them and read off the part you want.

Take x² − x − 12 > 0. The roots of (x − 4)(x + 3) = 0 are x = −3 and x = 4. The parabola opens upwards, so it is below the axis between the roots and above it outside them. Figure 3(a) shows it. The answer is x < −3 or x > 4.

Now 2x² + 5x ≤ 3. Rearrange: 2x² + 5x − 3 ≤ 0, which factorises as (2x − 1)(x + 3) ≤ 0, with roots −3 and ½. We want on or below the axis, which is between the roots: −3 ≤ x ≤ ½, as Figure 3(b) shows.

Three rules keep these answers right.

Inside or outside. For a > 0, "< 0" is always between the roots (one interval) and "> 0" is always outside them (two pieces). For a < 0 it is the other way round. The sketch tells you which, so always draw it, even small.

Two pieces are joined by "or". Write x < −3 or x > 4. Never write −3 > x > 4, which says x is less than −3 and greater than 4 at once, and is true for no number.

Strict or not. A strict inequality (> or <) excludes the roots; a non-strict one (≥ or ≤) includes them. The difference is one mark.

On Paper 2 the roots may be ugly. For 0.4x² − 3x + 2 < 0, the GDC gives the zeros of y = 0.4x² − 3x + 2 as x = 0.740 and x = 6.76, and the parabola opens upwards, so the answer is 0.740 < x < 6.76. The method is the same; only the root-finding changes.

8Questions with an unknown coefficient

The most characteristic Paper 1 question in this subtopic gives a quadratic with a letter in it and asks for the values of that letter which produce a certain kind of root. The discriminant answers it, and the answer often needs the inequality skills of section 7.

Example. For the equation kx² + 4x + (k − 3) = 0, find the values of k for which it has (a) two equal real roots, (b) two distinct real roots, (c) no real roots.

Write a, b and c down, then form Δ and simplify it.

(a) Equal roots: Δ = 0, so k = 4 or k = −1.

(b) Two distinct real roots: Δ > 0, so −4(k − 4)(k + 1) > 0, which means (k − 4)(k + 1) < 0 (dividing by −4 flips the sign). That is between the roots: −1 < k < 4. But k = 0 is in that interval, and at k = 0 the equation is 4x − 3 = 0, which is linear with only one root. So the answer is −1 < k < 4, k ≠ 0.

(c) No real roots: Δ < 0, so (k − 4)(k + 1) > 0, which is outside the roots: k < −1 or k > 4.

Figure 4 shows both sides of the story: in (a), three members of the family of parabolas, one for each case; in (b), the discriminant itself drawn as a function of k, positive only between −1 and 4.

Tangents are the same question. A line meets a curve where their equations are equal. If that gives a quadratic, the line is a tangent, touching the curve at one point, exactly when Δ = 0. For which gradients m is y = mx + 1 a tangent to y = x² + 4?

Figure 5 shows the two tangents from the point (0, 1), and the line with m = 4, for which Δ = 4 > 0: it meets the curve twice, at x = 1 and x = 3.

9Quadratics in context

A ball is kicked upwards so that its height after t seconds is h(t) = −4.9t² + 18t + 2 metres. For how long is it at least 10 m above the ground?

This is a quadratic inequality: −4.9t² + 18t + 2 ≥ 10, or −4.9t² + 18t − 8 ≥ 0. On Paper 2, find where the graph of h meets the line h = 10 with the GDC (or use the formula):

Figure 6 shows the arc above the line. Notice the last line: subtract the unrounded values, not 3.16 − 0.517 = 2.643, which happens to round the same here but will not always. And in any context, check that every root is in the domain; a negative time is thrown away.