Quadratics

• The completed square gives the vertex and the range; the discriminant \(b^2 - 4ac\) — no square root — counts the real roots; every inequality is read from a sketch.
• \(a(x + p)^2 + q\) has its vertex at \((-p,\ q)\). The discriminant is \(b^2 - 4ac\): \(> 0\) two distinct, \(= 0\) repeated, \(< 0\) none. Outside the roots is two inequalities joined by or.

• 1.1.1 I can — carry out the process of completing the square for a quadratic polynomial ax² + bx + c and use a completed square form
• 1.1.2 I can — find the discriminant of a quadratic polynomial ax² + bx + c and use the discriminant
• 1.1.3 I can — solve quadratic equations, and quadratic inequalities, in one unknown
• 1.1.4 I can — solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic
• 1.1.5 I can — recognise and solve equations in x which are quadratic in some function of x

Accuracy for this chapter. Leave surd answers exact: \(3 \pm \sqrt{7}\), \(2 \pm \sqrt{7}\), \(3 \pm \tfrac{1}{2}\sqrt{26}\). Give decimals only when the question asks for them, and then to 3 significant figures, carrying the unrounded values (at least four significant figures) through the working. An answer taken from a calculator's equation solver with no working shown earns no marks: the substitution into the formula, or the factorised form, is the working.

• “The discriminant is \(\sqrt{b^2 - 4ac}\).” Correct The discriminant is \(b^2 - 4ac\) itself, with no square root. It can be negative, and its sign is the whole point: positive, zero or negative decides two distinct real roots, one repeated root, or no real roots.
• “\(x^2 - 4x - 12 > 0\), so \(-2 > x > 6\).” Correct No number is both less than \(-2\) and greater than \(6\), so that statement is empty. “Outside the roots” is two inequalities joined by or: \(x < -2\) or \(x > 6\). Only “between the roots” is one chained statement, \(-2 < x < 6\).
• “The vertex of \(y = 2(x - 3)^2 - 13\) is \((-3, -13)\).” Correct \((x - 3)^2\) is zero when \(x = 3\), so the vertex is \((3, -13)\). In \(a(x + p)^2 + q\) the vertex is \((-p,\ q)\): the sign of the \(x\)-coordinate flips, the sign of \(q\) does not.
• “Real roots, so \(b^2 - 4ac > 0\).” Correct “Real roots” includes the repeated root, so it means \(b^2 - 4ac \geqslant 0\). The strict \(b^2 - 4ac > 0\) is for two distinct real roots. The question's wording decides whether your answer for \(k\) has \(<\) or \(\leqslant\).
• “\(2x^2 - 12x + 5 = 2(x - 3)^2 - 9 + 5\).” Correct When you take out \(a = 2\), the \(-9\) is inside the bracket that is multiplied by 2: \(2[(x - 3)^2 - 9] + 5 = 2(x - 3)^2 - 13\).
• “Divide \(x^2 > 5x\) by \(x\) to get \(x > 5\).” Correct \(x\) might be negative, which reverses the inequality, or zero. Never divide an inequality by an expression in \(x\): rearrange to \(x^2 - 5x > 0\), sketch, and read \(x < 0\) or \(x > 5\).
• “\(\sqrt{x} = \tfrac{1}{3}\) or \(-\tfrac{1}{2}\), so \(x = \tfrac{1}{9}\) or \(\tfrac{1}{4}\).” Correct \(\sqrt{x}\) is never negative, so \(\sqrt{x} = -\tfrac{1}{2}\) is rejected before you square anything. State the reason for every rejected value.
• “Substitute the \(x\)-values back into the circle to find \(y\).” Correct Back-substitute into the linear equation. The quadratic gives two \(y\)-values for each \(x\), and half of the resulting pairs are not solutions.
• “\(x^2 - 6x + 5 = (x - 3)^2 + 5\).” Repair Halve the 6 and subtract the square of the half: \((x - 3)^2 - 9 + 5 = (x - 3)^2 - 4\). Expanding \((x - 3)^2 + 5\) gives \(x^2 - 6x + 14\), which shows the error.
• “\(2x^2 - 12x + 5 = 2(x - 3)^2 - 9 + 5\).” Repair The \(-9\) is inside the bracket that is multiplied by 2: \(2[(x - 3)^2 - 9] + 5 = 2(x - 3)^2 - 13\).
• “The vertex of \(y = 2(x - 3)^2 - 13\) is \((-3, -13)\).” Repair \((x - 3)^2\) is zero at \(x = 3\), so the vertex is \((3, -13)\).
• “The discriminant is \(\sqrt{b^2 - 4ac}\).” Repair The discriminant is \(b^2 - 4ac\) itself, without the root. It can be negative.
• “Real roots, so \(b^2 - 4ac > 0\).” Repair Real roots means \(b^2 - 4ac \geqslant 0\); the strict inequality is for distinct real roots. Read the question's word.
• “\(x^2 - 4x - 12 > 0\), so \(x > 6\) or \(x > -2\).” Repair Sketch it: \(\cup\)-shaped through \(-2\) and \(6\), positive outside the roots, so \(x < -2\) or \(x > 6\).
• “\(-2 > x > 6\).” Repair No number satisfies that statement. Outside the roots is two inequalities joined by or: \(x < -2\) or \(x > 6\).
• “\(x^2 > 9\), so \(x > 3\).” Repair \(x^2 - 9 > 0\), critical values \(\pm 3\), \(\cup\)-shaped, so \(x < -3\) or \(x > 3\). \(x = -4\) satisfies \(x^2 > 9\) and the wrong answer loses it.
• “Divide \(2x^2 + 5x \leqslant 3x\) by \(x\).” Repair \(x\) may be negative, which reverses the inequality, or zero. Rearrange to zero, factorise and sketch: \(2x^2 + 2x \leqslant 0 \Rightarrow 2x(x + 1) \leqslant 0 \Rightarrow -1 \leqslant x \leqslant 0\). Dividing by \(x\) would have given \(x \leqslant -1\), which is wrong: \(x = -2\) gives \(8 - 10 \leqslant -6\), false.
• “Substituting \(x = 3\) into \(x^2 + y^2 = 25\) gives \(y = \pm 4\), so \((3, 4)\) and \((3, -4)\) are both solutions.” Repair Back-substitute into the linear equation \(x + y + 1 = 0\): \(x = 3\) gives only \(y = -4\). \((3, 4)\) is on the circle but not on the line.
• “\(6x + \sqrt{x} - 1 = 0\) gives \(\sqrt{x} = \tfrac{1}{3}\) or \(-\tfrac{1}{2}\), so \(x = \tfrac{1}{9}\) or \(\tfrac{1}{4}\).” Repair \(\sqrt{x}\) is never negative; reject \(-\tfrac{1}{2}\). \(x = \tfrac{1}{4}\) gives \(1.5 + 0.5 - 1 = 1 \neq 0\).
• “\(x^4 - 5x^2 + 4 = 0 \Rightarrow x^2 = 1\) or \(4 \Rightarrow x = 1\) or \(2\).” Repair \(x^2 = 1\) gives \(x = \pm 1\) and \(x^2 = 4\) gives \(x = \pm 2\): four roots.
• “\(k^2 + 4 = 0\), so \(k = \pm 2\).” Repair \(k^2 = -4\) has no real solution. “There is no real value of \(k\)” is a legitimate answer, with the working as its evidence.
• A “show that \(k \leqslant \tfrac{1}{4}\)” answer that begins “if \(k \leqslant \tfrac{1}{4}\) then…”. Repair Begin from the given condition (real roots, so \(b^2 - 4ac \geqslant 0\)) and derive \(k \leqslant \tfrac{1}{4}\), as in worked example 7. Working backwards from the result shows nothing.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. When you reach chapter 3, re-answer worked example 3 with a circle in place of the parabola: the tangency condition is still a repeated root. When you reach chapter 5, finish retrieval question 16 by solving \(\tan x = 1\) and \(\tan x = 2\) in an interval. When you reach chapter 7, find the vertex of \(y = 2x^2 - 12x + 5\) by differentiating and confirm it is \((3, -13)\). Recalling a method inside a new problem is worth more than another pass over this chapter on its own: Paper 1 is assumed knowledge for every other paper, and an individual examination question may involve ideas and methods from more than one section of that paper’s content, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 1 content, examined in Paper 1. Paper 1 (Pure Mathematics 1) is compulsory for both the AS Level and the A Level: it is 60% of the AS Level and 30% of the A Level, and its content is assumed knowledge for every other paper. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• Section 1.1 is written into structured questions in a few recognisable forms: “express in the form \(a(x + p)^2 + q\)” followed by “hence state the vertex” or “the range”; “find the set of values of \(k\) for which…”, where the discriminant produces an inequality in \(k\); a line and a curve, where you find the points of intersection or the value of a constant that makes the line a tangent; and an equation that is a quadratic in \(x^2\), \(\sqrt{x}\) or a trigonometric function. The reasoning sits in the set-up line: the condition on \(b^2 - 4ac\), the substitution \(u = \ldots\), the sketch that fixes which side of the critical values you want.
• MF19 prints the quadratic formula, \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), and nothing else from this section. Completing the square, the discriminant conditions, the vertex \((-p,\ q)\) and the inequality method are not in the booklet: you must know them. The discriminant is the part under the square root in the printed formula, which is one way to remember it.
• Answers that come out as surds stay exact: \(3 \pm \sqrt{7}\), \(3 \pm \tfrac{1}{2}\sqrt{26}\), \(2 - \sqrt{7} < x < 2 + \sqrt{7}\). Give decimals only when the question asks, then to 3 significant figures, and never round \(\sqrt{7}\) half-way through. Write the substitution into the formula, the discriminant condition and the factorised form before any number, because an answer read off a calculator equation solver with no working earns nothing.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.