HookWhy the Highway Code hides a quadratic in every stopping distance
The Highway Code prints a table every UK learner has to memorise: at \(30\) mph a car needs about \(23\) metres to stop, but at \(70\) mph it needs about \(96\) metres. That is not a bit more than double for a bit more than double the speed — it is more than four times the distance. The reason is buried in the arithmetic, and it is the reason this section exists. A stopping distance is not one thing but two added together: the thinking distance you travel while your brain reacts, plus the braking distance you travel once the brakes bite.
Write the standard model with the speed \(v\) in miles per hour and the distance \(d\) in feet and it comes out as \(d=v+\dfrac{v^2}{20}\). The thinking part, \(v\), is a straight line — react-time is roughly constant, so it grows in step with speed. The braking part, \(\dfrac{v^2}{20}\), is a quadratic — it depends on the square of the speed, which is why doubling your speed quadruples how far you skid. Ask 'what speed makes the stopping distance exactly \(315\) feet?' and you are solving a quadratic. Ask 'what speeds keep me stopping within a limit?' and you are solving an inequality. This section is the full toolkit for both: linear equations, quadratics by three different methods, pairs of simultaneous equations, iteration for the equations that refuse to factorise, turning a worded situation into algebra, and handling inequalities and their solution sets.
MechanismLinear equations — including the unknown on both sides
A linear equation has the unknown to the first power only, no \(x^2\). You solve it by doing the same inverse operation to both sides until \(x\) stands alone, undoing the operations in reverse order. When the unknown appears on both sides, the extra first move is to gather all the \(x\)-terms on one side and all the numbers on the other, usually by subtracting the smaller \(x\)-term from both sides so the coefficient stays positive.
Brackets are cleared by expanding first; fractions are cleared by multiplying every term by the denominator. AQA also asks for the graphical method: the solution of \(5x-3=2x+9\) is the \(x\)-coordinate where the lines \(y=5x-3\) and \(y=2x+9\) cross, and more generally an equation \(f(x)=g(x)\) is solved by reading off where the two graphs intersect. That link — an algebraic solution is a graphical crossing point — is the thread running through this whole section.
Solve \(5(x-1)=2x+7\).
Expand the bracket first: \(5x-5=2x+7\). Now gather the unknowns by subtracting \(2x\) from both sides: \(3x-5=7\). Add \(5\) to both sides: \(3x=12\). Divide by \(3\): \(x=4\). Check by substituting back into the original: left side \(5(4-1)=15\), right side \(2(4)+7=15\) ✓. Graphically, \(x=4\) is exactly where the lines \(y=5(x-1)\) and \(y=2x+7\) would cross — the algebra and the picture agree.
MechanismQuadratics three ways — factorising, completing the square, the formula
A quadratic \(ax^2+bx+c=0\) can have two solutions, one, or none, and Higher expects all three methods. Factorising is quickest when it works: write the quadratic as two brackets and set each to zero, because if a product is zero then one of its factors must be zero. Completing the square rewrites it as \((x+p)^2+q=0\) and solves by taking square roots — the method that always works and gives exact surd answers. The quadratic formula, \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\), is the universal tool: it solves any quadratic, especially the many that will not factorise.
The quantity under the root, \(b^2-4ac\), is the discriminant, and it tells you the number of solutions before you finish: if \(b^2-4ac\gt 0\) there are two distinct real roots, if it equals \(0\) there is one repeated root, and if \(b^2-4ac\lt 0\) there are no real roots because you cannot square-root a negative. Reading the discriminant first is a mark-earning habit and stops you hunting for roots that do not exist.
Solve \(2x^2-5x-4=0\), giving answers to \(2\) decimal places.
This does not factorise with whole numbers, so use the formula with \(a=2,\ b=-5,\ c=-4\). First the discriminant: \(b^2-4ac=(-5)^2-4(2)(-4)=25+32=57\). Since \(57\gt 0\), expect two real roots. Now substitute: \(x=\dfrac{-(-5)\pm\sqrt{57}}{2(2)}=\dfrac{5\pm\sqrt{57}}{4}\). With \(\sqrt{57}\approx 7.550\): \(x=\dfrac{5+7.550}{4}\approx 3.14\) or \(x=\dfrac{5-7.550}{4}\approx -0.64\). The two brackets around \((-5)\) and \((-4)\) are what keep the signs correct — losing the double negative in \(-4(2)(-4)\) is the most common formula error.
MechanismSimultaneous equations — two linear, or one linear and one quadratic
Two equations in two unknowns pin down a point (or points) where both are true at once — graphically, where the two curves cross. For two linear equations, use elimination: scale one or both equations so a variable has matching coefficients, then add or subtract to remove it, solve for the survivor, and back-substitute for the other.
When one equation is linear and one is quadratic, elimination will not work; use substitution instead. Rearrange the linear equation to make one variable the subject, substitute it into the quadratic, and you are left with a single quadratic to solve by the methods above. Because a line can cut a parabola or circle in two places, expect up to two solution pairs — and each \(x\) must be paired with its own \(y\), never mixed and matched.
Solve \(y=x+1\) and \(x^2+y^2=25\) simultaneously.
Substitute the linear equation into the quadratic: \(x^2+(x+1)^2=25\). Expand: \(x^2+x^2+2x+1=25\), so \(2x^2+2x-24=0\). Divide through by \(2\): \(x^2+x-12=0\), which factorises to \((x+4)(x-3)=0\), giving \(x=-4\) or \(x=3\). Now pair each with its \(y\) using \(y=x+1\): when \(x=3,\ y=4\); when \(x=-4,\ y=-3\). The two solutions are \((3,4)\) and \((-4,-3)\) — the two points where the line crosses the circle of radius \(5\). Reporting \(x=3,\ y=-3\) by muddling the pairs is the classic dropped mark.
ModelIteration — cracking the equations that will not factorise
Plenty of equations have no neat algebraic solution at all — cubics like \(x^3-x-3=0\) are typical. Iteration finds a solution to as many decimal places as you like by repeated approximation, which is exactly how calculators and engineering software actually solve equations behind the scenes. First you show a root exists in an interval by a sign change: if \(f(a)\) and \(f(b)\) have opposite signs, a continuous curve must cross zero somewhere between them.
You then rearrange the equation into the form \(x=g(x)\) and feed each answer back in as the next input, using \(x_{n+1}=g(x_n)\) from a sensible starting value. If the sequence settles down, the number it homes in on is the root. The care points are to keep every digit on your calculator between steps (rounding early corrupts the answer) and to write each iterate to enough decimal places to show it converging.
Show that \(x^3-x-3=0\) has a root between \(1\) and \(2\), then find it to \(3\) decimal places using \(x_{n+1}=\sqrt[3]{x_n+3}\).
Let \(f(x)=x^3-x-3\). Then \(f(1)=1-1-3=-3\) and \(f(2)=8-2-3=3\). The sign changes from negative to positive, so a root lies between \(1\) and \(2\). Now iterate from \(x_0=1\): \(x_1=\sqrt[3]{1+3}=1.587\); \(x_2=\sqrt[3]{4.587}=1.662\); \(x_3=\sqrt[3]{4.662}=1.670\); \(x_4=\sqrt[3]{4.670}=1.671\). The values have stopped changing in the third decimal place, so the root is \(x\approx 1.671\). The rearrangement to \(x=\sqrt[3]{x+3}\) came from adding \(x+3\) to both sides of \(x^3=x+3\) and cube-rooting — a different rearrangement can converge faster, or fail to converge at all.
CaseFrom words to algebra — forming and interpreting an equation
The hardest marks on this paper are not the solving but the setting up: turning a sentence into an equation, solving it, then interpreting the answer back in the real context. The routine is to name the unknown with a letter, translate each clause into algebra, form the equation, solve it, and finally reject any solution that makes no physical sense — a negative length, a negative speed, a fractional person.
The stopping-distance model from the hook is the perfect worked case, because it forces every step: you build a quadratic from a real formula, solve it, and read a number that turns out to be a familiar speed limit. Interpreting the answer — saying what the number means — is worth its own mark, and it is the one candidates most often forget to write.
On a motorway a driver must be able to stop within \(315\) feet (about \(96\) m) of clear road. Using \(d=v+\dfrac{v^2}{20}\), with \(d\) in feet and \(v\) in mph, find the maximum safe speed.
Set the stopping distance equal to the limit: \(v+\dfrac{v^2}{20}=315\). Multiply every term by \(20\) to clear the fraction: \(20v+v^2=6300\). Rearrange into standard form: \(v^2+20v-6300=0\). Factorise — two numbers multiplying to \(-6300\) and adding to \(20\) are \(90\) and \(-70\): \((v+90)(v-70)=0\), so \(v=-90\) or \(v=70\). A speed cannot be negative, so reject \(-90\). The maximum safe speed is \(v=70\) mph — and interpreting that, it is no accident: the national motorway limit is set exactly where a car needs about \(96\) m to stop.
MechanismInequalities — solution sets on a line, in set notation and on a graph
An inequality is solved almost like an equation, with one rule that must never be broken: multiplying or dividing by a negative number reverses the sign, so \(-2x\lt 6\) becomes \(x\gt -3\). A linear inequality gives a range, shown on a number line with an open circle for a strict \(\lt\) or \(\gt\) and a filled circle for an inclusive \(\leq\) or \(\geq\), or written in set notation such as \(\{x : x\gt 2\}\).
A quadratic inequality needs its shape. Solve the matching equation to find the critical values, sketch the parabola, then read off which region satisfies the inequality: below the axis (between the roots) for \(\lt 0\), or above the axis (outside the roots) for \(\gt 0\). Inequalities in two variables describe a region of the plane: draw the boundary line — dashed for strict, solid for inclusive — and shade the side that satisfies it, testing a point like the origin to decide which side. The answer to an inequality is always a set of values, never a single number.
For a car to stop within \(175\) feet, its speed must satisfy \(v+\dfrac{v^2}{20}\leq 175\). Find the safe speeds.
Multiply by \(20\): \(20v+v^2\leq 3500\), so \(v^2+20v-3500\leq 0\). Solve the equation \(v^2+20v-3500=0\) for the critical values: two numbers multiplying to \(-3500\) and adding to \(20\) are \(70\) and \(-50\), so \((v+70)(v-50)=0\) gives \(v=-70\) or \(v=50\). The parabola opens upwards, so it is at or below zero between its roots: \(-70\leq v\leq 50\). A speed cannot be negative, so the physical answer is \(0\leq v\leq 50\), or in set notation \(\{v : 0\leq v\leq 50\}\). Interpreted: to pull up inside \(175\) feet you must be doing no more than \(50\) mph.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This topic is spread across the non-calculator (Paper 1) and calculator (Papers 2 and 3) papers, and it is marked method-first, so lay out every stage. For quadratics, state which method you are using and, when you reach for the formula, write the discriminant \(b^2-4ac\) as a separate line — it earns credit and confirms how many roots to expect. Keep negatives in brackets when you substitute into the formula; the lost double-negative in \(-4ac\) is the most frequent error in the whole section.
On 'form and solve' questions, define your variable in a sentence before you write any algebra, and finish with an interpretation line that rejects impossible solutions and states what the valid answer means in context — that final sentence is a mark in itself. For iteration, show the sign-change check first, then list every iterate to the required decimal places so the convergence is visible, and never round mid-sequence. For inequalities, remember to flip the sign when dividing by a negative, sketch the parabola for any quadratic inequality to read the correct region, and give your final answer as a set or a range on a number line — a single value can never be the full answer to an inequality. Where a question offers a graph, use intersections as solutions: the meeting point of two lines solves a linear equation, and the crossings of a line and a curve solve a linear-and-quadratic pair.