Quadratic Equations & Graphs

Method	Use it when	The signal in the question
Factorisationⓘ Recap	the expression factorises over the integers	no accuracy is stated, or the answers are "nice"
Completing the square	the coefficient of `x`² is 1, or you also need the turning point	"by completing the square", or "express in the form (`x` + `r`)² + `u`"
Quadratic formula	always works — and it is the only route when `a` ≠ 1 and it will not factorise	"correct to 2 decimal places" / "3 significant figures"
Graphical method	you are given, or asked to draw, the graph of `y` = `ax`² + `bx` + `c`	"draw the graph … hence solve"

Factorisation — the recap ⓘ Recap

Everything in this chapter rests on one idea you already have:

Zero Product Principle. If P and Q are factors of an algebraic expression such that PQ = 0, then P = 0 or Q = 0.

To factorise x² + bx + c, look for two numbers whose product is c and whose sum is b.

Walkthrough 1 — solve by factorisation Basic

Solve the equation x² − 2x − 15 = 0.

Two numbers with product −15 and sum −2: −5 and +3.

Factors first: it is the fastest route, and it gives exact answers. The pair you want multiplies to c and adds to b.
(x − 5)(x + 3) = 0

Write the factorised form as a line of its own. This line carries the method mark, and the roots come out of it.
x − 5 = 0 or x + 3 = 0

The Zero Product Principle, written out. This is the line that turns one equation into two, one per bracket.
x = 5 or x = −3

Give both roots. A quadratic that factorises has exact roots, so 5 and −3 are the finished answers; "5.00" would claim a rounding that never happened.

Where this comes from

So the moment a quadratic is written as a product of two brackets equal to zero, the two answers fall out: the Zero Product Principle says one bracket or the other is zero, and each of those is a linear equation. Not every quadratic factorises. x² + 4x + 3 = 0 does; x² + 4x − 3 = 0 does not, because no pair of integers multiplies to −3 and adds to 4. That is the whole reason the rest of this tab exists: completing the square and the formula work whether or not the brackets are there.

Past-paper questions

G3 N2018 · Paper 1 · Q3 2 marks

One solution of the equation kx² + (k + 1)x − 4 = 0 is x = −2.Find

the value of k, [1]
the second possible value of x. [1]

Worked solution

(a) x = −2 satisfies the equation, so substitute it: k(−2)² + (k + 1)(−2) − 4 = 0 4k − 2k − 2 − 4 = 0, so 2k = 6 and k = 3. (b) With k = 3 the equation becomes 3x² + 4x − 4 = 0. (3x − 2)(x + 2) = 0, so x =

\frac{2}{3}

or x = −2.

−2 is the root the question already gave, so the second value is x = $\frac{2}{3}$ .

Check: 3 ${(\frac{2}{3})}^{2}$ + 4 $(\frac{2}{3})$ − 4 = $\frac{4}{3}$ + $\frac{8}{3}$ − 4 = 0.

Why this works. With k unknown, the given solution x = −2 is the way in: substituting it turns the stem into an equation for k. That substitution line is the mark for (a), brackets and all: (−2)² = +4, and −2 multiplies the whole bracket (k + 1). Part (b) is an ordinary factorisation, and −2 is spent, so the second value is $\frac{2}{3}$ .

G3 N2019 · Paper 1 · Q8 3 marks

Use factorisation to solve the equation.

3p² + p − 10 = 0

[3]

Worked solution

The coefficient of p² is not 1, so split the middle term. Two numbers with product 3 × (−10) = −30 and sum +1: +6 and −5.

3p² + 6p − 5p − 10 = 0

3p(p + 2) − 5(p + 2) = 0

(p + 2)(3p − 5) = 0

p + 2 = 0 or 3p − 5 = 0

p = −2 or p = $\frac{5}{3}$

Check: 3(−2)² + (−2) − 10 = 12 − 12 = 0, and 3 ${(\frac{5}{3})}^{2}$ + $\frac{5}{3}$ − 10 = $\frac{25}{3}$ + $\frac{5}{3}$ − 10 = 0.

Why this works. "Use factorisation" names the method, so the three marks are on the route: split, group, bracket, roots. With a = 3 the pair of numbers multiplies to ac = −30, not to c = −10. After the split the two grouped brackets come out identical; if they do not, swap the pair over, and finish exactly at $\frac{5}{3}$ .

G3 N2021 · Paper 1 · Q21(b) 3 marks

Using factorisation, solve 8x² − 6x − 9 = 0. [3]

Worked solution

Two numbers with product 8 × (−9) = −72 and sum −6: −12 and +6.

8x² − 12x + 6x − 9 = 0

4x(2x − 3) + 3(2x − 3) = 0

(4x + 3)(2x − 3) = 0

x = − $\frac{3}{4}$ or x = $\frac{3}{2}$

Check: the two roots multiply to $- \frac{9}{8}$ , which is $\frac{c}{a} = \frac{−9}{8}$ , and they add to $\frac{3}{4}$ , which is $- \frac{b}{a} = \frac{6}{8}$ .

Why this works. With a = 8 and c = −9 both composite, guessing brackets by inspection means sifting many sign-and-factor combinations, so split the middle term instead: one pair, found from −72 and −6. Keep each sign attached to its number when grouping — +6x − 9 factorises as +3(2x − 3), which is what makes the two brackets match. Both roots are exact fractions.

Completing the square

To complete the square for x² + bx, add

{(\frac{b}{2})}^{2}

, so that the completed square is

{(x + \frac{b}{2})}^{2}

.

Complete the square only for y = x² + px + q — the syllabus form, with coefficient of x² equal to 1. If it is not 1, divide by a first, or (quicker in an exam) use the formula.

Walkthrough 2 — complete the square for an expression Basic

Express x² − 7x + 5 in the form (x + r)² + u.

The coefficient of x is −7, and half of it is $- \frac{7}{2}$ .

Completing the square starts with half the coefficient of x. That half is r, so the bracket can be written down before any other arithmetic.
$x^{2} - 7 x + 5 = [x^{2} - 7 x + {(- \frac{7}{2})}^{2}] - {(- \frac{7}{2})}^{2} + 5$

Add the square and subtract it again, so the expression is worth exactly what it was worth before. The + 5 is left alone — the constant term stays outside the square.
$= {(x - \frac{7}{2})}^{2} - \frac{49}{4} + 5$

The bracket closes into a perfect square, and (−7/2)² = 49/4 — square the whole thing, minus sign included, so the subtracted term is positive 49/4.
$= {(x - \frac{7}{2})}^{2} - \frac{29}{4}$

−49/4 + 20/4 = −29/4. "In the form (x + r)² + u" means exact values of r and u, so they are left as fractions rather than −3.5 and −7.25.

Walkthrough 3 — solve by completing the square Intermediate

Solve x² + 10x − 3 = 0 by completing the square, giving your answers correct to 2 decimal places.

x² + 10x = 3

Move the constant across so the left side is bare x² + 10x — that is the only shape a square can be completed on. The equation stays balanced because the same 3 was added to both sides.
x² + 10x + 5² = 3 + 5²

Half of 10 is 5, so add 5² to both sides. Added to one side only, it would turn this into a different equation.
(x + 5)² = 28

The left side is now a perfect square and 3 + 25 = 28. This is the line the method mark is attached to.
$x + 5 = \pm \sqrt{28}$

Take square roots of both sides and keep the ±. The negative root is the second of the two answers.
$x = - 5 \pm \sqrt{28}$

Keep the surd exact until the last line. Rounding √28 here and then subtracting can push the final answers outside the accepted tolerance.
x = 0.29 or x = −10.29 (2 d.p.)

√28 = 5.2915…, so −5 + 5.2915 = 0.2915 and −5 − 5.2915 = −10.2915. Both roots, rounded once, to the accuracy asked for — "2 decimal places" said from the start that they would not be whole numbers.

Where this comes from

Because you cannot simply add something to an expression and leave it unchanged, you subtract it again:

x^{2} + b x = {(x + \frac{b}{2})}^{2} - {(\frac{b}{2})}^{2}

Two equivalent routes. Either add and subtract (b/2)² on the left only, as Walkthrough 2 does, or move the constant across and add (b/2)² to both sides, as Walkthrough 3 does. The second is usually cleaner. Either way the equation stays balanced at every line.

Past-paper questions

G3 N2021 · Paper 1 · Q16(a) 1 mark

The formula s = 3.5 (t − 2)² can be used to calculate the distance, s metres, travelled by a cyclist in a time of t seconds, where t > 2.
Find t when s = 14. [1]

Worked solution

3.5(t − 2)² = 14

(t − 2)² = 4

t − 2 = 2 or t − 2 = −2

t = 4 or t = 0

t = 0 is rejected because the formula only holds for t > 2, so t = 4.

Check: 3.5(4 − 2)² = 3.5 × 4 = 14 m.

Why this works. The formula arrives already in completed-square form, so divide by 3.5 and square-root both sides: expanding into 3.5t² − 14t + 14 = 14 gets there too, but turns one mark into four lines. Keep the ±, because a squared bracket equal to 4 has two roots. The printed condition t > 2 is what disposes of t = 0.

G3 N2022 · Paper 2 · Q1(c) 4 marks

Solve the equation x² + 9x − 16 = 0 by completing the square.
Give your solutions correct to two decimal places. [4]

Worked solution

Half the coefficient of x is $\frac{9}{2}$ , so move the constant across and add ${(\frac{9}{2})}^{2}$ to both sides:

x² + 9x = 16

$x^{2} + 9 x + \frac{81}{4} = 16 + \frac{81}{4}$

${(x + \frac{9}{2})}^{2} = \frac{145}{4} = 36.25$

$x + 4.5 = \pm \sqrt{36.25} = \pm 6.020797…$

x = −4.5 + 6.020797… = 1.520797… or x = −4.5 − 6.020797… = −10.520797…

x = 1.52 or x = −10.52 (2 d.p.)

Why this works. "By completing the square" names the method, so the marks sit on the completing lines: $\frac{81}{4}$ added to both sides, the perfect square, the ±, the rounding. The number to get right is ${(\frac{9}{2})}^{2} = 20.25$ , and moving the constant across first puts 16 + 20.25 = 36.25 on the right. Keep $\sqrt{36.25}$ to six figures until the last line.

G3 N2024 · Paper 1 · Q25 3 marks

Solve the equation x² − 12x + 17 = 0 by completing the square.
Give your solutions correct to 2 decimal places. [3]

Worked solution

Half the coefficient of x is −6, so add and subtract (−6)² = 36:

(x − 6)² − 36 + 17 = 0

(x − 6)² = 19

$x - 6 = \pm \sqrt{19} = \pm 4.358898…$

x = 6 + 4.358898… = 10.358898… or x = 6 − 4.358898… = 1.641101…

x = 10.36 or x = 1.64 (2 d.p.)

Check: the two roots add to 12, which is $- \frac{b}{a}$ , and multiply to 36 − 19 = 17, which is $\frac{c}{a}$ .

Why this works. An even, negative coefficient of x is friendly: half of −12 is whole, so the bracket is (x − 6) and the square to subtract is 36. The sign inside the bracket copies that of the x term. Then −36 + 17 = −19 moves across as (x − 6)² = +19: a positive right-hand side, so two real roots exist, to 2 d.p.

The quadratic formula

One formula solves every quadratic equation:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

The part under the square root, b² − 4ac, decides how many real solutions there are:

`b`² − 4`ac`	Number of real solutions	What the graph does
> 0	two different solutions	cuts the `x`-axis twice
= 0	one solution (a repeated root)	touches the `x`-axis once
< 0	no real solutions	never meets the `x`-axis

Write a, b and c down before touching the calculator, signs included. In 3x² − 4x − 5 = 0, c is −5, not 5, and −4(3)(−5) = +60. The signs are where this method is won.

Walkthrough 4 — solve with the formula Intermediate

Solve 2x² − 7x − 3 = 0, giving your answers correct to 3 significant figures.

a = 2, b = −7, c = −3

The coefficient of x² is not 1, so completing the square is not the syllabus route here — the formula is. Note the signs travel with the numbers: c = −3.
$x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 (2) (- 3)}}{2 (2)}$

Substitute with every negative inside its own bracket. Two signs decide this line: (−7)² = +49, and −4(2)(−3) = +24.
$= \frac{7 \pm \sqrt{73}}{4}$

b² − 4ac = 49 + 24 = 73. It is positive, so there are two different real roots — a one-line sanity check worth making before you round.
x = 3.89 or x = −0.386 (3 s.f.)

√73 = 8.5440…, so x = 15.544 ÷ 4 = 3.886 and x = −1.544 ÷ 4 = −0.3860. 3 significant figures is the O-Level default whenever the question does not say otherwise.

Walkthrough 5 — rearrange first, then solve Advanced

Solve (x + 4)(x − 2) = 5x − 3, giving your answers correct to 2 decimal places.

x² + 2x − 8 = 5x − 3

Expand the brackets first. Neither the formula nor completing the square means anything until the equation has been rearranged to = 0.
x² − 3x − 5 = 0

Take everything to the left: 2x − 5x = −3x and −8 + 3 = −5. Read a, b, c off this line, not off the un-rearranged equation.
b² − 4ac = (−3)² − 4(1)(−5) = 9 + 20 = 29

A separate discriminant line is cheap insurance — if it had come out negative you would know at once that a sign had gone astray.
$x = \frac{3 \pm \sqrt{29}}{2}$

−b = −(−3) = +3. This line is the exact answer; everything after it is only rounding.
x = 4.19 or x = −1.19 (2 d.p.)

√29 = 5.3851…, so x = 8.3851 ÷ 2 = 4.1926 and x = −2.3851 ÷ 2 = −1.1926. Round once, at the end, and give both.

Where this comes from

The formula is what completing the square gives when it is done once and for all on ax² + bx + c = 0, with a, b and c left as letters instead of numbers. Because the letters stand for any numbers at all, the result solves every quadratic equation, whether or not it factorises — which is why it is the one method that always finishes.

Check yourself

Solve x² + 3x − 28 = 0 by factorisation.
Answer
Two numbers with product −28 and sum +3: +7 and −4.
(x + 7)(x − 4) = 0
x + 7 = 0 or x − 4 = 0
x = −7 or x = 4 (exact — do not round).
Solve (2y − 3)² = 20, giving your answers correct to 2 decimal places.
Answer
2y − 3 = ±√20
2y = 3 ± 4.4721…
y = (3 ± 4.4721) ÷ 2
y = 3.74 or y = −0.74 (2 d.p.). Take the square root of both sides before expanding — expanding first only makes more work.
Express x² + 9x − 1 in the form (x + r)² + u.
Answer
Half of 9 is 9/2, so
x² + 9x − 1 = (x + 9/2)² − 81/4 − 1
= (x + 9/2)² − 85/4. Keep r and u exact.
Solve x² − 5x − 2 = 0 by completing the square, giving your answers correct to 2 decimal places.
Answer
x² − 5x = 2
x² − 5x + (5/2)² = 2 + 25/4
(x − 5/2)² = 33/4
x − 5/2 = ±√(33/4) = ±2.8722…
x = 2.5 ± 2.8722
x = 5.37 or x = −0.37 (2 d.p.).
Solve 3x² = 4x + 6, giving your answers correct to 3 significant figures.
Answer
Rearrange first: 3x² − 4x − 6 = 0, so a = 3, b = −4, c = −6.
b² − 4ac = 16 + 72 = 88
x = (4 ± √88) ÷ 6 = (4 ± 9.3808…) ÷ 6
x = 2.23 or x = −0.897 (3 s.f.).
Why does "give your answers correct to 2 decimal places" tell you the equation will not factorise?
Answer
A quadratic that factorises over the integers has exact roots, which need no rounding. An instruction to round is a signal that the roots are irrational, so go straight to the formula (or completing the square) instead of hunting for factors that are not there.

Fractional equations that reduce to quadratics

An equation containing one or more algebraic fractions is a fractional equation. There is no way to solve one directly — so you turn it into an equivalent equation you can already solve.

Step	What you do	Why it works
0	note which values make a denominator zero	those values can never be solutions, whatever the algebra says
1	multiply both sides by the LCM of the denominators	every fraction cancels, leaving a polynomial equation
2	expand and rearrange to `ax`² + `bx` + `c` = 0	this is the quadratic the question was hiding
3	solve by factorisation or the formula	the methods of tab 2.1, unchanged
4	reject any root found in step 0, giving the reason	it makes the original equation undefined

Multiply every term, on both sides. Write the multiplication out in full as the first line: it costs one line and protects several marks.

Walkthrough 6 — one fraction Basic

Solve

\frac{3}{x + 1} = 2 x + 5

, giving your answers correct to 3 significant figures.

3 = (2x + 5)(x + 1)

Multiply both sides by (x + 1), the only denominator. Clearing the fraction in a single line is what turns this into something you already know how to solve.
3 = 2x² + 7x + 5

Expand carefully: 2x × x = 2x², then 2x × 1 + 5 × x = 7x, then 5 × 1 = 5.
2x² + 7x + 2 = 0

Rearrange to = 0. "Reduced to a quadratic equation" — the exact phrase in the syllabus outcome — has now happened, and it is worth a mark on its own.
b² − 4ac = 7² − 4(2)(2) = 33

b² − 4ac earns a line of its own: it says how many real roots there are, and whether they are exact. 33 is positive but not a perfect square, so there are two real roots and neither is whole.
$x = \frac{- 7 \pm \sqrt{33}}{4}$ = −0.314 or −3.19 (3 s.f.)

√33 = 5.7445…, giving −1.2555 ÷ 4 and −12.7445 ÷ 4 — the formula finishes what factors could not. Neither root makes x + 1 = 0, so both stand, to the 3 significant figures asked for.

Walkthrough 7 — two fractions, using the LCM Intermediate

Solve

\frac{4}{x + 1} + \frac{x - 2}{3 - x} = 3

, giving your answers correct to 3 significant figures.

LCM of the denominators = (x + 1)(3 − x)

Two different denominators with no common factor, so their LCM is simply their product. Write it down before you multiply — it is the plan for the whole question.
4(3 − x) + (x − 2)(x + 1) = 3(x + 1)(3 − x)

Every term on both sides is multiplied by the LCM. Each fraction loses its own denominator; the 3 on the right gains the whole product.
12 − 4x + x² − x − 2 = 3(−x² + 2x + 3)

Expand both sides. On the right, (x + 1)(3 − x) = −x² + 2x + 3 — the reversed bracket 3 − x is where signs go wrong.
x² − 5x + 10 = −3x² + 6x + 9

Tidy each side before moving anything across. Two short lines beat one long one for keeping signs straight.
4x² − 11x + 1 = 0

Bring everything to the left. This is the quadratic the whole question was building towards, and it carries a method mark of its own.
$x = \frac{11 \pm \sqrt{105}}{8}$ = 2.66 or 0.0941 (3 s.f.)

b² − 4ac = 121 − 16 = 105 and √105 = 10.2469… Neither answer makes x + 1 or 3 − x zero, so both stand. Note 0.0941 has three significant figures: leading zeros do not count.

Walkthrough 8 — the root you must reject Advanced

Solve

\frac{x + 4}{x - 2} = \frac{6}{x - 2} + x

.

x − 2 ≠ 0, so x ≠ 2.

Write down the forbidden value before you do any algebra. A denominator can never be zero, and this whole question is built on that fact.
x + 4 = 6 + x(x − 2)

Multiply through by (x − 2). Both fractions share that denominator, so it clears in one step — but the lone x on the right must be multiplied too.
x + 4 = 6 + x² − 2x

Expand. The + 6 stays outside the bracket — it was never part of the product.
x² − 3x + 2 = 0

Collect on the right-hand side and swap: −2x − x = −3x and 6 − 4 = +2.
(x − 1)(x − 2) = 0, so x = 1 or x = 2

It factorises, so both roots are exact: no formula, no rounding. Write the factorised line down before the roots — it is the method mark.
x = 2 is rejected because it makes x − 2 = 0, so x = 1.

The algebra is not the last word in a fractional equation: substituting x = 2 into the original would divide by zero, so it is not a solution of that equation. Give the reason beside the rejection.

More detail

The term that gets missed is the one on the right-hand side: it carries no fraction of its own, so it looks as though it can be left alone. If it is not multiplied by the LCM as well, everything from line two onwards belongs to a different equation, and the quadratic that comes out is not the one the question asked about.

Check yourself

Solve 8/x = 2x + 3, giving your answers correct to 3 significant figures.
Answer
x ≠ 0.
8 = x(2x + 3) = 2x² + 3x
2x² + 3x − 8 = 0
b² − 4ac = 9 + 64 = 73
x = (−3 ± √73) ÷ 4 = (−3 ± 8.5440) ÷ 4
x = 1.39 or x = −2.89 (3 s.f.); neither is 0, so both are valid.
Solve (x + 2)/(4 − x) = x.
Answer
x ≠ 4.
x + 2 = x(4 − x) = 4x − x²
x² − 3x + 2 = 0
(x − 1)(x − 2) = 0
x = 1 or x = 2 — neither equals 4, so both are kept, and both are exact.
Solve 1/(x + 3) + 2/(x − 1) = 1, giving your answers correct to 3 significant figures.
Answer
x ≠ −3 and x ≠ 1. Multiply throughout by (x + 3)(x − 1):
(x − 1) + 2(x + 3) = (x + 3)(x − 1)
3x + 5 = x² + 2x − 3
x² − x − 8 = 0
b² − 4ac = 1 + 32 = 33
x = (1 ± √33) ÷ 2 = (1 ± 5.7446) ÷ 2
x = 3.37 or x = −2.37 (3 s.f.).
Solve 5/(x − 1) − 3/(x − 1)² = 1, giving your answers correct to 3 significant figures.
Answer
x ≠ 1. The LCM is (x − 1)²:
5(x − 1) − 3 = (x − 1)²
5x − 8 = x² − 2x + 1
x² − 7x + 9 = 0
b² − 4ac = 49 − 36 = 13
x = (7 ± √13) ÷ 2 = (7 ± 3.6056) ÷ 2
x = 5.30 or x = 1.70 (3 s.f.); neither is 1, so both stand.
A student solves 3/(x − 4) − 1/x = 12/(x(x − 4)) and writes "x = 4". Explain why that is not a solution, and state what the answer should be.
Answer
Multiplying by x(x − 4) gives 3x − (x − 4) = 12, so 2x + 4 = 12 and x = 4. But x = 4 makes the denominator x − 4 equal to zero, so it cannot be substituted into the original equation. It must be rejected — and since it was the only value produced, the equation has no solution.

Past-paper questions

G3 N2015 · Paper 2 · Q1(c), (d) 7 marks

Solve the equation $\frac{9}{x - 4} = 2 x - 1$ . [3]
1. Express x² − 9x + 17 in the form (x + a)² + b. [1]
2. Hence solve the equation x² − 9x + 17 = 0, giving your answers correct to two decimal places. [3]

Worked solution

(a) x = 4 would make the denominator zero, so it can never be a solution. Multiplying both sides by (x − 4):

9 = (2x − 1)(x − 4)

9 = 2x² − 8x − x + 4

2x² − 9x − 5 = 0

(2x + 1)(x − 5) = 0

x = −0.5 or x = 5 (neither is 4, so both stand)

(b)(i) Half the coefficient of x is −4.5, so

x² − 9x + 17 = (x − 4.5)² − 4.5² + 17 = (x − 4.5)² − 3.25, so a = −4.5 and b = −3.25.

(b)(ii) Using the form just found — that is what Hence asks for:

(x − 4.5)² − 3.25 = 0, so (x − 4.5)² = 3.25

$x - 4.5 = \pm \sqrt{3.25} = \pm 1.802775…$

x = 4.5 + 1.802775… = 6.302775… or x = 4.5 − 1.802775… = 2.697224…

x = 6.30 or x = 2.70 (2 d.p.)

Why this works. The first line carries the mark in (a): multiplying by (x − 4) clears the fraction on the left and multiplies the whole of 2x − 1 on the right, brackets included. In (b), Hence ties the parts: the (x − 4.5)² − 3.25 earned in (i) is what (ii) reuses. Write 2.70 with the trailing zero: "2.7" does not show two decimal places.

G3 N2017 · Paper 2 · Q1(e) 3 marks

Solve the equation $\frac{2}{x - 2} + \frac{3}{x + 4} = 1$ . [3]

Worked solution

x = 2 and x = −4 make a denominator zero, so neither can be a solution. Multiply every term by the LCM (x − 2)(x + 4):

2(x + 4) + 3(x − 2) = (x − 2)(x + 4)

2x + 8 + 3x − 6 = x² + 2x − 8

5x + 2 = x² + 2x − 8

x² − 3x − 10 = 0

(x − 5)(x + 2) = 0

x = 5 or x = −2

Check x = −2: $\frac{2}{−4} + \frac{3}{2} = −0.5 + 1.5 = 1$ .

Why this works. Two fractions make the LCM the product of the denominators, and all three terms are multiplied by it — the 1 on the right included, which is what makes the equation quadratic. Expand (x − 2)(x + 4) before collecting, so the −8 lands on the correct side. Both roots survive the check against x = 2 and x = −4, so nothing is rejected.

G3 N2018 · Paper 2 · Q1(c) 3 marks

Solve the equation $\frac{20}{x + 1} = 2 x + 5$ [3]

Worked solution

x = −1 makes the denominator zero, so it cannot be a solution. Multiplying both sides by (x + 1):

20 = (2x + 5)(x + 1)

20 = 2x² + 2x + 5x + 5

2x² + 7x − 15 = 0

(2x − 3)(x + 5) = 0

x = 1.5 or x = −5

Check x = −5: $\frac{20}{−4} = −5$ and 2(−5) + 5 = −5.

Why this works. A single fraction is the friendly case: one multiplication clears it, no LCM needed, and the line worth the method mark is 20 = (2x + 5)(x + 1), written before any expanding. The negative root stands: a fractional equation loses a root only when that root makes a denominator zero, and −5 does not. There is no length or speed here to be negative.

G3 N2019 · Paper 2 · Q1(d) 4 marks

Solve $\frac{1}{x - 3} + \frac{6}{x - 1} = 2$ . [4]

Worked solution

x = 3 and x = 1 make a denominator zero. Multiply every term by (x − 3)(x − 1):

(x − 1) + 6(x − 3) = 2(x − 3)(x − 1)

x − 1 + 6x − 18 = 2(x² − 4x + 3)

7x − 19 = 2x² − 8x + 6

2x² − 15x + 25 = 0

(2x − 5)(x − 5) = 0

x = 2.5 or x = 5

Check x = 2.5: $\frac{1}{−0.5} + \frac{6}{1.5} = −2 + 4 = 2$ .

Why this works. The plain 2 on the right is what makes this four marks: times the LCM it becomes a bracketed product to expand. Expand 2(x − 3)(x − 1) as 2x² − 8x + 6, since the 2 multiplies every term. Collect on the side that keeps the x² coefficient positive: 2x² − 15x + 25 factorises on sight.

G3 N2021 · Paper 2 · Q1(c) 5 marks

Solve the equation $\frac{3 x}{x - 4} + \frac{5}{2 x + 1} = 1$ .
Give your solutions correct to 2 decimal places. [5]

Worked solution

x = 4 and x = −0.5 make a denominator zero. Multiply every term by (x − 4)(2x + 1):

3x(2x + 1) + 5(x − 4) = (x − 4)(2x + 1)

6x² + 3x + 5x − 20 = 2x² − 7x − 4

4x² + 15x − 16 = 0

b² − 4ac = 15² − 4(4)(−16) = 225 + 256 = 481, which is not a perfect square, so the quadratic does not factorise — use the formula:

$x = \frac{−15 \pm \sqrt{481}}{2 (4)} = \frac{−15 \pm 21.931712…}{8}$

x = 0.866464… or x = −4.616464…

x = 0.87 or x = −4.62 (2 d.p.); neither is 4 or −0.5, so both stand.

Why this works. Five marks buys a longer road: clear, expand three products, collect, solve. "Correct to 2 decimal places" says in advance the quadratic will not factorise, so 4x² + 15x − 16 = 0 not coming apart is confirmation. Two signs decide it: (x − 4)(2x + 1) expands to 2x² − 7x − 4, and −4(4)(−16) is +256.

G3 N2023 · Paper 2 · Q2(d) 5 marks

Solve the equation $\frac{x}{2 x - 1} - \frac{6}{4 - x} = 3$ .
Give your solutions correct to two decimal places. [5]

Worked solution

x = 0.5 and x = 4 make a denominator zero. Multiply every term by (2x − 1)(4 − x):

x(4 − x) − 6(2x − 1) = 3(2x − 1)(4 − x)

4x − x² − 12x + 6 = 3(−2x² + 9x − 4)

−x² − 8x + 6 = −6x² + 27x − 12

5x² − 35x + 18 = 0

b² − 4ac = (−35)² − 4(5)(18) = 1225 − 360 = 865, so

$x = \frac{35 \pm \sqrt{865}}{2 (5)} = \frac{35 \pm 29.410882…}{10}$

x = 6.441088… or x = 0.558911…

x = 6.44 or x = 0.56 (2 d.p.); neither is 0.5 or 4, so both stand.

Why this works. The minus between the two fractions is the whole difficulty: it is carried onto both terms of 6(2x − 1), giving −12x + 6. The root 0.56 sits close to the excluded value x = 0.5, which is why the excluded values are written down at the start. Keep six figures of $\sqrt{865}$ through the division by 10.

What a quadratic graph looks like

The graph of y = ax² + bx + c is a parabola. Its shape is decided by one thing only — the sign of a — and it is always symmetrical about a vertical line through its turning point (the maximum or minimum point).

a > 0 → the curve opens upwards and the turning point is a minimum.
a < 0 → the curve opens downwards and the turning point is a maximum.

Fig. 2.1

The four critical points of a quadratic graph: the y-intercept, the x-intercepts (if there are any), and the turning point — plus the line of symmetry through it.

Drawing a graph is not the same as sketching one

	Drawing (plotting)	Sketching
Paper	graph paper	plain or foolscap paper
Scale on the axes	yes, always	yes — without one the curve comes out distorted
Points needed	about 7 or more, at equally spaced `x`-values	only the critical points: `y`-intercept, `x`-intercepts (if any), turning point
Can you read off other values?	Yes — that is the point of plotting it	No — an un-plotted point on a sketch means nothing

Past-paper questions

G3 N2018 · Paper 1 · Q21 4 marks

The points (−2, 11) and (4, −1) lie on the curve given by the equation y = ax² + bx + 3.
Use an algebraic method to find the values of a and b. [4]

Worked solution

A point lies on the curve, so its coordinates satisfy the equation.

When x = −2 and y = 11: 11 = a(−2)² + b(−2) + 3, so 4a − 2b = 8, that is 2a − b = 4 …(1)

When x = 4 and y = −1: −1 = a(4)² + b(4) + 3, so 16a + 4b = −4, that is 4a + b = −1 …(2)

(1) + (2): 6a = 3, so a = 0.5.

Substituting into (1): 2(0.5) − b = 4, so b = −3.

Check (4, −1): 0.5(16) − 3(4) + 3 = 8 − 12 + 3 = −1.

Why this works. "Lies on the curve" means the coordinates satisfy the equation; two unknown coefficients need two points, and the printed + 3 is the y-intercept, handed over so that two are enough. "Use an algebraic method" means both substitution lines appear on the page. The square kills the minus while the linear term keeps it: a(−2)² = +4a, b(−2) = −2b.

Sketching from a given form ✕ Not in G2

Two forms hand you the critical points almost for free. Recognising which one you have is most of the question.

Factorised form y = ±(x − a)(x − b)

the x-intercepts are a and b;
the line of symmetry is $x = \frac{a + b}{2}$ ;
the x-coordinate of the turning point is $\frac{a + b}{2}$ — halfway between the intercepts.

Completed-square form y = ±(x − p)² + q

the turning point is (p, q);
the line of symmetry is x = p.

The sketching checklist. (1) shape from the sign of a; (2) y-intercept; (3) x-intercepts, if any; (4) turning point; (5) scaled axes, a cross on each critical point, one smooth symmetrical curve, the equation beside it. The marks are on the labels.

Walkthrough 9 — sketch y = (x − a)(x − b) Basic

Sketch the graph of y = (x + 2)(x − 4), showing clearly the coordinates of the points where the graph crosses the axes and of the turning point.

Expanding gives y = x² − 2x − 8, so the coefficient of x² is +1: the curve is U-shaped, with a minimum point.

Decide the shape first, so the sketch starts from a U and not an ∩. Only the sign of the x² term is needed, not the full expansion.
When x = 0, y = (0 + 2)(0 − 4) = −8 → (0, −8)

The y-intercept is the cheapest critical point there is: put x = 0. Write it as a coordinate, because that is what the question asked for.
When y = 0, (x + 2)(x − 4) = 0, so x = −2 or x = 4 → (−2, 0) and (4, 0)

In factorised form the x-intercepts can be read straight off: the graph of y = (x − a)(x − b) cuts the axis at a and b. No expanding, no formula.
x-coordinate of the turning point $= \frac{- 2 + 4}{2} = 1$

The curve is symmetrical, so the turning point sits exactly halfway between the two x-intercepts — the average of a and b.
y = (1 + 2)(1 − 4) = (3)(−3) = −9, so the minimum point is (1, −9)

Substitute back into the factorised form — far less arithmetic than using the expanded one, and far less to get wrong.
Fig. 2.2

Axes with a scale, crosses on the four critical points, coordinates written in, and the equation labelled on the curve.

The sketch itself carries the remaining marks, and they are all for labels: a scale on both axes, a cross on each critical point, its coordinates, and the equation written beside the curve. Draw one smooth curve, symmetrical about x = 1.

Walkthrough 10 — sketch y = (x − p)² + q Basic

Sketch the graph of y = (x + 1)² − 9, showing clearly the coordinates of the points where the graph crosses the axes and of the minimum point.

Comparing with y = (x − p)² + q: p = −1 and q = −9, so the minimum point is (−1, −9).

Completed-square form hands you the turning point as (p, q). Mind the sign: (x + 1)² is (x − (−1))², so p = −1, not +1.
When x = 0, y = (0 + 1)² − 9 = −8 → (0, −8)

y-intercept as usual: put x = 0 into the form you were given. No expanding first.
When y = 0, (x + 1)² = 9

Leave the square whole. Isolating it finishes the job in one square-root step — the same move as the last stage of completing the square.
x + 1 = ±3, so x = 2 or x = −4 → (2, 0) and (−4, 0)

Keep the ± and both x-intercepts appear; with only one the curve could not be symmetrical. Two crossings is what q = −9 promised: a U-shaped curve whose minimum sits below the axis cuts it twice.
Fig. 2.3

A U-shaped curve through the four critical points, symmetrical about x = −1.

The remaining marks are in the labelling: scaled axes, a cross and its coordinates on each of the four critical points, and y = (x + 1)² − 9 on the curve itself. Draw one smooth curve, symmetrical about x = −1.

Walkthrough 11 — complete the square, then sketch a ∩-shaped curve Intermediate

(i) Express y = −x² − 4x + 5 in the form y = −(x + p)² + q. (ii) Hence sketch the graph, showing the coordinates of the axis crossings and of the turning point.

y = −(x² + 4x) + 5

Factor −1 out of the x² and x terms only, so the bracket begins with a positive x². The + 5 stays outside — it never joins the square.
= −(x² + 4x + 2² − 2²) + 5

Half of 4 is 2, so add and subtract 2² inside the bracket. Both are inside, so the value of the expression has not changed.
= −[(x + 2)² − 4] + 5 = −(x + 2)² + 9

The minus sign outside turns −4 into +4, and 4 + 5 = 9. Most of part (i)'s marks sit on that one sign step.
The coefficient of x² is negative, so the curve is ∩-shaped, with a maximum point at (−2, 9).

Rewriting −(x + 2)² + 9 as −(x − (−2))² + 9 puts the turning point at (−2, 9). Say the word "maximum" — naming the kind of turning point is part of the answer.
When x = 0, y = −(0 + 2)² + 9 = 5 → (0, 5)

Check it against the original: −0² − 4(0) + 5 = 5. Two forms agreeing is a free confirmation that part (i) was right.
When y = 0, (x + 2)² = 9, so x = 1 or x = −5 → (1, 0) and (−5, 0)

With y = 0, (x + 2)² = 9, the squared term changing sign as it crosses, so x + 2 = ±3. Two crossings is what q = 9 predicted: a maximum above the axis has to come back down through it.
Fig. 2.4

A ∩-shaped curve with maximum point (−2, 9), symmetrical about x = −2.

Part (ii) is marked on the labels: scaled axes, a cross and its coordinates on each of the four critical points, and y = −(x + 2)² + 9 written on the curve. Draw one smooth ∩-shape, symmetrical about x = −2.

Walkthrough 12 — a curve that never meets the x-axis Advanced

(i) Express y = x² + 2x + 5 in the form y = (x + p)² + q. (ii) Hence sketch the graph, and explain why the curve never meets the x-axis.

y = x² + 2x + 1² − 1² + 5

Half of 2 is 1: add and subtract 1². The coefficient of x² is already 1, so nothing has to be factored out first.
= (x + 1)² + 4

−1 + 5 = 4, so p = 1 and q = 4. Two lines is all part (i) needs.
The coefficient of x² is positive, so the curve is U-shaped with a minimum point at (−1, 4).

Read the turning point straight off the completed square: (x + 1)² is (x − (−1))², so it is (−1, 4); the + in front of the square makes it a minimum.
When x = 0, y = (0 + 1)² + 4 = 5 → (0, 5)

Still worth finding: with no x-intercepts, the y-intercept is one of only two critical points you have.
When y = 0: (x + 1)² = −4, which has no real solution.

A square can never be negative, so no real value of x satisfies this. That single line is the explanation the question is asking for.
Since (x + 1)² ≥ 0 for all real x, y ≥ 4 > 0, so the curve lies entirely above the x-axis.

The same fact in the picture's language: the lowest point of the curve is 4 above the axis, so it cannot reach it. That is what q = 4 being positive said in part (i); either wording earns the mark.
Fig. 2.5

Minimum point (−1, 4), y-intercept (0, 5), and its mirror image (−2, 5) — three points are enough.

With only two critical points, manufacture a third by reflecting the y-intercept in the line of symmetry x = −1: (0, 5) becomes (−2, 5). Without it you cannot control the shape of the left-hand arm.

More detail

Every quadratic function can be written in completed-square form; only those whose graphs actually cut the x-axis can be written in factorised form. So when a question gives you a quadratic with no obvious factors, complete the square: that route is open whatever the curve does, which is why Walkthrough 12 can sketch a curve with no x-intercepts at all.

Solving a quadratic equation graphically

The equation of the x-axis is y = 0. So the solutions of ax² + bx + c = 0 are exactly the x-coordinates of the points where the curve y = ax² + bx + c cuts the x-axis — its x-intercepts.

How accurately can you read a graph? Only to about half of the smallest square — roughly ±0.1 on a normal scale. So quote graphical answers to 1 decimal place. Writing 3 s.f. claims a precision the graph does not have.

Walkthrough 13 — draw the graph, hence solve Intermediate

The variables x and y are connected by y = x² − 3x − 3. (i) Complete the table of values. (ii) Draw the graph for −2 ≤ x ≤ 5. (iii) Hence solve x² − 3x − 3 = 0.

x −2 −1 0 1 2 3 4 5

y 7 1 −3 −5 −5 −3 1 7

The table carries no marks of its own, but every value in it feeds the curve that does. Free check: the values are symmetrical (7, 1, −3, −5 | −5, −3, 1, 7), which is what a parabola must do.
Plot the eight points and join them with a single smooth curve — no ruled segments, no feathered double lines.

The drawing mark is for a smooth curve of the right shape through the plotted points — one continuous line, drawn in a single pass.
The solutions of x² − 3x − 3 = 0 are the x-coordinates of the points where the curve meets the line y = 0 (the x-axis).

Write this sentence down — it is the "hence" the question asked for, and it is the whole idea of the graphical method. Solving f(x) = 0 means finding where y = f(x) meets y = 0.
Fig. 2.6

Reading the two x-intercepts: x = −0.8 and x = 3.8.

From the graph, x = −0.8 or x = 3.8.

Read each crossing to the nearest half-square and give 1 decimal place. (The formula gives −0.7912… and 3.7912…, so both readings are well inside the tolerance a graphical answer is allowed.)

`x`	−2	−1	0	1	2	3	4	5
`y`	7	1	−3	−5	−5	−3	1	7

More detail

The number of x-intercepts is the discriminant, drawn. Two crossings means b² − 4ac > 0; one touch means it is 0; no crossing means it is negative and the equation has no real solutions. The algebra of the discriminant and the picture of the curve are the same statement.

Check yourself

Write down the coordinates of the turning point and the equation of the line of symmetry of y = (x − 4)² + 3, and state whether the turning point is a maximum or a minimum. ✕ Not in G2
Answer
Comparing with y = (x − p)² + q gives p = 4 and q = 3.
Turning point (4, 3); line of symmetry x = 4; it is a minimum, because the coefficient of x² is positive.
Sketch the graph of y = (x − 1)(x + 5), showing clearly the coordinates of the axis crossings and of the turning point. ✕ Not in G2
Answer
Coefficient of x² is positive → U-shaped, minimum.
When x = 0: y = (−1)(5) = −5 → (0, −5).
When y = 0: x = 1 or x = −5 → (1, 0) and (−5, 0).
Turning point: x = (1 + (−5)) ÷ 2 = −2, and y = (−2 − 1)(−2 + 5) = (−3)(3) = −9 → (−2, −9).
Draw scaled axes, cross the four points, join with one smooth curve symmetrical about x = −2, and label it y = (x − 1)(x + 5).
Express y = x² − 6x + 4 in the form y = (x − p)² + q, and hence write down the coordinates of the minimum point. ✕ Not in G2
Answer
Half of −6 is −3:
y = x² − 6x + 3² − 3² + 4 = (x − 3)² − 5
Minimum point (3, −5).
Explain, without drawing anything, how you can tell that the graph of y = −(x − 2)² − 1 never crosses the x-axis. ✕ Not in G2
Answer
(x − 2)² ≥ 0 for all real x, so −(x − 2)² ≤ 0 and therefore y ≤ −1. The greatest value of y is −1, which is below the x-axis, so the whole curve lies below the axis and never crosses it. (Equivalently: y = 0 would need (x − 2)² = −1, which has no real solution.)
A student has drawn the graph of y = x² − 4x + 1 for 0 ≤ x ≤ 5 and is asked to solve x² − 4x + 1 = 0. Where on the graph are the answers, and to what accuracy should they be given?
Answer
The answers are the x-coordinates of the two points where the curve cuts the x-axis (the line y = 0). Read each to the nearest half small square and give 1 decimal place: x ≈ 0.3 or 3.7. (The exact roots are 2 ± √3 = 0.2679… and 3.7320…)
Give two differences between drawing a graph and sketching one.
Answer
Any two of: drawing uses graph paper and a sketch uses plain paper; drawing needs about 7 or more plotted points at equal intervals, while a sketch needs only the critical points (axis crossings and the turning point); values at un-plotted points can be read off a drawn graph but not off a sketch. Both still need a scale on the axes, or the curve comes out distorted.

Formulating a quadratic from a word problem

Paper 2 ends with a question set in a real-world context, and quadratics are one of its favourite tools. The algebra is the easy half — the marks are in setting the equation up and in answering the question that was asked.

The four moves. 1. Say what x is. 2. Write every other quantity in terms of x. 3. Form the equation from the sentence not yet used. 4. Solve, reject the impossible root with a reason, then answer in context, with units.

"Show that it simplifies to …" means the printed equation has to appear on your page; write (shown) beside it. If your algebra does not reach it, go back and find where it went, rather than adjusting the last line.

Carry extra figures between the parts of a question. Keep 5 or 6 figures in the calculator from one part to the next, and round only in the final line of each part.

Reject with a reason. Lengths, speeds, times, masses and numbers of objects cannot be negative — and a number of people or cakes must also be a whole number. Write the reason out: "rejected because a length cannot be negative".

Walkthrough 14 — Pythagoras, a scale, and a rejected root Intermediate

On a map drawn to a scale of 1 : 2000, a plot of land is a right-angled triangle with sides x cm, (x + 4) cm, and hypotenuse (2x + 1) cm. (i) Show that 2x² − 4x − 15 = 0. (ii) Solve the equation. (iii) Explain why one solution must be rejected. (iv) Find the actual length of fencing needed to go right round the plot, in metres.

(2x + 1)² = x² + (x + 4)² (Pythagoras' Theorem)

The hypotenuse is the longest side, (2x + 1), so it goes alone on one side of the equation. Naming the theorem is worth the method mark on its own.
4x² + 4x + 1 = x² + x² + 8x + 16

Expand both squares in full: (2x + 1)² = 4x² + 4x + 1, not 4x² + 1 — a bracket squares as a whole, not term by term.
2x² − 4x − 15 = 0 (shown)

4x² − 2x² = 2x², 4x − 8x = −4x, 1 − 16 = −15. The printed equation has now appeared exactly, so write (shown) beside it.
$x = \frac{4 \pm \sqrt{136}}{4}$

a = 2, b = −4, c = −15, so b² − 4ac = 16 + 120 = 136 — not a perfect square, so the formula it is.
x = 3.91548 or x = −1.91548

Keep five or six figures. Part (iv) is built on this number, and rounding to 3 s.f. now would visibly shift the final answer.
x = −1.91548 is rejected because a length cannot be negative, so x = 3.92 (3 s.f.).

Give the reason: part (iii) asks for exactly that, and "reject the negative value" on its own does not say why.
Perimeter on the map = x + (x + 4) + (2x + 1) = 4x + 5 = 4(3.91548) + 5 = 20.6619 cm

Simplify the expression before substituting — 4x + 5 is one multiplication instead of three additions — and use the unrounded x.
Actual length = 20.6619 × 2000 = 41 323.8 cm = 413 m (3 s.f.)

A scale of 1 : 2000 means 1 cm on the map is 2000 cm on the ground. Convert to metres at the very end (÷ 100), and round only once — and give the unit, because the question asked for metres.

Walkthrough 15 — speed, distance and time Advanced

A van drives 200 km from a depot to a warehouse at an average speed of x km/h. A car covers the same distance at an average speed 15 km/h greater and takes 45 minutes less. (i) Show that x² + 15x − 4000 = 0. (ii) Solve it, giving both answers correct to 3 decimal places. (iii) Find the time taken by the car, to the nearest minute.

Time taken by the van $= \frac{200}{x}$ h

time = distance ÷ speed. Write each quantity in terms of x before trying to form an equation — that is move 2 of the four.
Time taken by the car $= \frac{200}{x + 15}$ h

"15 km/h greater" makes the car's speed (x + 15) — the whole bracket goes in the denominator. 200/x + 15 would be a different quantity entirely.
45 min $= \frac{45}{60} = \frac{3}{4}$ h

The speeds are in km/h, so the time difference goes in hours. Minutes and hours cannot share one equation.
$\frac{200}{x} - \frac{200}{x + 15} = \frac{3}{4}$

The van is slower, so it takes longer: van time − car time is the positive difference. Subtracting the other way round gives a negative left-hand side and an equation that goes nowhere.
800(x + 15) − 800x = 3x(x + 15)

Multiply throughout by the LCM 4x(x + 15). All three terms must be multiplied — including the 3/4 on the right.
12 000 = 3x² + 45x, so x² + 15x − 4000 = 0 (shown)

800x cancels on the left, then divide throughout by 3 to land on exactly the printed equation.
$x = \frac{- 15 \pm \sqrt{16225}}{2}$ = 56.189 or −71.189 (3 d.p.)

b² − 4ac = 225 + 16 000 = 16 225. The question asks for 3 decimal places precisely because part (iii) needs that extra accuracy.
x > 0, so the van's speed is 56.189 km/h; time by car = 200 ÷ (56.189 + 15) = 2.8094 h = 169 min (2 h 49 min)

Reject the negative root — a speed cannot be negative — and say so. Then convert the decimal part of the hour: 0.8094 × 60 = 48.6 min, so 2.81 h reads as 2 h 49 min.

Walkthrough 16 — modelling with a quadratic function Intermediate

A ball is thrown upwards from a balcony. Its height h metres above the ground, t seconds after it is released, is modelled by h = −t² + 6t + 2. Find the greatest height the ball reaches, and how much higher that is than the point from which it was thrown.

When t = 0, h = −0 + 0 + 2 = 2, so the ball is released 2 m above the ground.

t = 0 is the moment of release, so the constant term is the starting height — the h-intercept of the model. The question needs it for the second half of the answer.
h = −(t² − 6t) + 2

The greatest height is the maximum point, and completed-square form hands it over. Factor −1 out of the t² and t terms first.
= −(t² − 6t + 3² − 3²) + 2

Half of −6 is −3, so add and subtract 3² inside the bracket — exactly as in tab 2.1, just with t instead of x.
= −(t − 3)² + 9 + 2 = −(t − 3)² + 11

The minus sign outside turns −9 into +9, and 9 + 2 = 11. A sign slip here would turn a maximum of 11 into one of −7.
Since −(t − 3)² ≤ 0, the greatest value of h is 11 m, reached when t = 3 s.

A square is never negative, so h is largest when the square is zero. Give the time as well — "when" is normally worth a mark of its own.
Difference in height = 11 − 2 = 9 m

Answer the question actually asked: the rise above the release point, not the greatest height. The last line of the question is the one to re-read before writing the answer.

More detail

Move 1 is written in words — "let the speed of the van be x km/h" — so that the marker knows what the letter stands for. Move 3 usually uses the sentence with "more than", "less than" or a total in it: that is the one comparison the earlier moves have not spent, and it is what makes an equation rather than an expression.

A "show that" hands you the answer to that part so that the next part is still worth attempting. That is why the printed equation has to appear on the page rather than be assumed, and why a mismatch is a signal to go back.

On accuracy, K310 says: "In questions which explicitly require an answer to be shown to be correct to a specific accuracy, the answer must be first shown to a higher degree of accuracy." That is the rule behind carrying 5 or 6 figures from one part of a question to the next.

Check yourself

A rectangular poster has a perimeter of 90 cm and an area of 464 cm². Taking the breadth to be x cm, show that x² − 45x + 464 = 0, and find the dimensions of the poster.
Answer
Perimeter 90 cm → length + breadth = 45, so the length is (45 − x) cm.
x(45 − x) = 464
45x − x² = 464
x² − 45x + 464 = 0 (shown)
b² − 4ac = 2025 − 1856 = 169, so x = (45 ± 13) ÷ 2 = 29 or 16.
Either root works — they are the two sides — so the poster is 16 cm by 29 cm.
Mei cycles 30 km at an average speed of x km/h. Raj cycles the same 30 km at a speed 5 km/h greater and arrives half an hour earlier. Show that x² + 5x − 300 = 0, and find Mei's speed.
Answer
Mei's time = 30/x h; Raj's time = 30/(x + 5) h.
30/x − 30/(x + 5) = 1/2
Multiply by 2x(x + 5): 60(x + 5) − 60x = x(x + 5)
300 = x² + 5x
x² + 5x − 300 = 0 (shown)
(x + 20)(x − 15) = 0, so x = 15 or x = −20.
x = −20 is rejected because a speed cannot be negative, so Mei's speed is 15 km/h. (Check: 2 h and 1.5 h — half an hour apart.)
Two consecutive positive odd numbers have a product of 483. Form a quadratic equation and find the two numbers.
Answer
Let the smaller number be x; the next odd number is x + 2.
x(x + 2) = 483
x² + 2x − 483 = 0
b² − 4ac = 4 + 1932 = 1936 = 44², so x = (−2 ± 44) ÷ 2 = 21 or −23.
x = −23 is rejected because the numbers are positive, so they are 21 and 23.
A stone is thrown from the top of a cliff. Its height above sea level, h metres, t seconds after release, is modelled by h = −t² + 4t + 21. Find (i) the greatest height it reaches, and (ii) the time at which it hits the water.
Answer
(i) h = −(t² − 4t) + 21 = −(t − 2)² + 4 + 21 = −(t − 2)² + 25.
Greatest height 25 m, when t = 2 s.
(ii) It hits the water when h = 0: t² − 4t − 21 = 0, so (t − 7)(t + 3) = 0 and t = 7 or t = −3.
t = −3 is rejected because time cannot be negative, so the stone hits the water after 7 s.
Printer A prints 60 pages every x minutes. Printer B takes 3 minutes longer than A to print 60 pages. Used together they print 45 pages in one minute. Show that 3x² + x − 12 = 0, and find how long Printer B takes to print 60 pages, correct to 3 significant figures.
Answer
In one minute A prints 60/x pages and B prints 60/(x + 3) pages.
60/x + 60/(x + 3) = 45
Multiply by x(x + 3): 60(x + 3) + 60x = 45x(x + 3)
120x + 180 = 45x² + 135x
45x² + 15x − 180 = 0, and dividing by 15, 3x² + x − 12 = 0 (shown)
b² − 4ac = 1 + 144 = 145, so x = (−1 ± √145) ÷ 6 = 1.8402… or −2.1735…
The negative root is rejected because a time cannot be negative, so x = 1.8402 min and Printer B takes 1.8402 + 3 = 4.84 min (3 s.f.).

Past-paper questions

G3 N2016 · Paper 2 · Q7(b) 10 marks

A small business makes mugs with designs painted on them.
David and Maryam paint designs on the mugs.

David takes x seconds to paint one mug.
Write an expression, in terms of x, for the number of mugs he paints in an hour. [1]
Maryam takes 50 seconds less than David to paint one mug.
Write an expression, in terms of x, for the number of mugs she paints in one hour. [1]
One morning, David and Maryam each work for 4 hours.
Altogether they paint a total of 68 mugs.
Write down an equation in x to represent this information and show that it reduces to
17x² − 8050x + 180 000 = 0. [3]
Solve the equation 17x² − 8050x + 180 000 = 0. [3]
Find the number of mugs Maryam paints in one hour. [2]

Worked solution

(a) One hour is 3600 seconds, so David paints $\frac{3600}{x}$ mugs in an hour.

(b) Maryam takes (x − 50) seconds per mug, so she paints $\frac{3600}{x - 50}$ mugs in an hour.

(c) Each works 4 hours, and the two totals add to 68:

$4 (\frac{3600}{x} + \frac{3600}{x - 50}) = 68$

Multiplying by x(x − 50):

14 400(x − 50) + 14 400x = 68x(x − 50)

28 800x − 720 000 = 68x² − 3400x

68x² − 32 200x + 720 000 = 0, and dividing by 4,

17x² − 8050x + 180 000 = 0 (shown)

(d) b² − 4ac = 8050² − 4(17)(180 000) = 64 802 500 − 12 240 000 = 52 562 500, and $\sqrt{52562500} = 7250$ exactly.

$x = \frac{8050 \pm 7250}{2 (17)} = \frac{15300}{34} or \frac{800}{34}$

x = 450 or x = 23.5 (3 s.f.), the second root being $\frac{400}{17} = 23.529411…$

(e) Maryam's time per mug is (x − 50) seconds, so x must be more than 50. That rejects x = 23.5, leaving x = 450.

Mugs Maryam paints in one hour $= \frac{3600}{450 - 50} = \frac{3600}{400}$ = 9.

Check: David paints 3600 ÷ 450 = 8 an hour, so in 4 hours the two of them paint 4(8) + 4(9) = 68 mugs.

Why this works. Everything runs on "per" turned upside down: seconds per mug inverts to mugs per second, and 3600 ÷ x makes that mugs per hour. In (c) the printed equation has to appear on the page — write (shown) — after multiplying all three terms by x(x − 50), the 68 included. In (e), 23.5 goes because x − 50 would be negative.

G3 N2017 · Paper 2 · Q7 11 marks

An open box is in the shape of a cuboid.
The height of the box is 5 cm less than its length.
The width of the box is twice its height.

Given that the length of the box is x cm, write down expressions, in terms of x, for the height and the width of the box. [2]
The external surface area of the open box is 48 cm².
Write down an equation to represent this information and show that it simplifies to
2x² − 15x + 13 = 0. [3]
Solve the equation 2x² − 15x + 13 = 0. [3]
Explain why one of the solutions in part (c) must be rejected as the length of the box. [1]
Calculate the volume of the box. [2]

Worked solution

(a) Height = (x − 5) cm; width = 2(x − 5) = (2x − 10) cm.

(b) The box is open, so there are five faces: the base, two sides (length × height) and two ends (width × height).

x(2x − 10) + 2x(x − 5) + 2(2x − 10)(x − 5) = 48

2x² − 10x + 2x² − 10x + 4x² − 40x + 100 = 48

8x² − 60x + 52 = 0, and dividing by 4,

2x² − 15x + 13 = 0 (shown)

(c) Two numbers with product 2 × 13 = 26 and sum −15: −13 and −2.

2x² − 13x − 2x + 13 = 0, so (2x − 13)(x − 1) = 0

x = 6.5 or x = 1

(d) If x = 1 then the height is 1 − 5 = −4 cm. A height cannot be negative, so x = 1 must be rejected.

(e) With x = 6.5: length 6.5 cm, height 6.5 − 5 = 1.5 cm, width 2(1.5) = 3 cm.

Volume = 6.5 × 3 × 1.5 = 29.25 cm³

Check on (b): the five faces are 6.5(3) + 2(6.5)(1.5) + 2(3)(1.5) = 19.5 + 19.5 + 9 = 48 cm².

Why this works. Everything turns on the word open: five faces, not six; the missing top is congruent to the base. Build the expressions in the order the sentences give them, and note the width is twice the height, not the length. In (d) the rejection names the quantity that goes negative: at x = 1 the length is fine, so the height rules the root out.

G3 N2018 · Paper 2 · Q6 11 marks

An empty water tank has a capacity of 3500 litres.
Pump A can fill the tank at a rate of x litres per minute.
Pump B can fill the tank at a rate of (x − 10) litres per minute.

Write down an expression, in terms of x, for the number of minutes it would take to fill the tank using pump A. [1]
Write down an expression, in terms of x, for the number of minutes it would take to fill the tank using pump B. [1]
It takes 21 minutes longer to fill the tank using pump B than it does using pump A.
Write down an equation to represent this information and show that it reduces to
3x² − 30x − 5000 = 0. [3]
Solve the equation 3x² − 30x − 5000 = 0, giving your solutions correct to two decimal places. [4]
Calculate how long it would take to fill the tank from empty using pump A and pump B together.
Give your answer in minutes and seconds, correct to the nearest ten seconds. [2]

Worked solution

(a) $\frac{3500}{x}$ minutes

(b) $\frac{3500}{x - 10}$ minutes

(c) Pump B's time is 21 minutes more than pump A's:

$\frac{3500}{x - 10} - \frac{3500}{x} = 21$

Multiplying by x(x − 10):

3500x − 3500(x − 10) = 21x(x − 10)

35 000 = 21x² − 210x

21x² − 210x − 35 000 = 0, and dividing by 7,

3x² − 30x − 5000 = 0 (shown)

(d) b² − 4ac = (−30)² − 4(3)(−5000) = 900 + 60 000 = 60 900

$x = \frac{30 \pm \sqrt{60900}}{2 (3)} = \frac{30 \pm 246.779253…}{6}$

x = 46.129875… or x = −36.129875…

x = 46.13 or x = −36.13 (2 d.p.)

(e) A rate cannot be negative, so x = 46.129875… Working together, the two pumps deliver x + (x − 10) = 82.259751… litres per minute.

Time $= \frac{3500}{82.259751…}$ = 42.548147… minutes = 42 minutes and 32.888… seconds

42 minutes 30 seconds (nearest ten seconds)

Why this works. Rates invert into times, so (a) and (b) are litres ÷ litres-per-minute, and the 21 minutes in (c) is a difference of times: write it as B − A, slower pump first. In (e) the rates add and the times do not, so add the rates and divide once, carrying the unrounded 46.129875… Then read the accuracy: 32.9 seconds to the nearest ten seconds is 30.

G3 N2024 · Paper 1 · Q21(a) 3 marks

12 24 8 21 28 17 2p 4p²

The list shows information about the number of text messages Li received each day for 8 days.

The mean number of text messages per day is 17.5 .

Show that p = 2.5 . [3]

Worked solution

The mean of the eight values is 17.5, so their total is 8 × 17.5 = 140.

12 + 24 + 8 + 21 + 28 + 17 = 110, so

110 + 2p + 4p² = 140

4p² + 2p − 30 = 0, and dividing by 2,

2p² + p − 15 = 0

(2p − 5)(p + 3) = 0

p = 2.5 or p = −3

p = −3 would make the seventh value 2p = −6, and a number of text messages cannot be negative, so it is rejected.

p = 2.5 (shown)

Check: the list becomes 12, 24, 8, 21, 28, 17, 5, 25, whose total is 140 and whose mean is 140 ÷ 8 = 17.5 .

Why this works. The statistics is one line — total = mean × number of values — and the rest is quadratic: with 4p² in the list, collect into 2p² + p − 15 = 0 and factorise. A "show that" whose algebra yields two answers asks you to dispose of one, for a mark. Name the value that goes wrong: 2p = −6 messages.

G3 N2024 · Paper 2 · Q6 11 marks

Nadia drives the first 40 km of a journey on an expressway and the remaining 20 km on local roads.
She drives on the expressway at an average speed of x km/h.
Her average speed on the local roads is 30 km/h slower than her average speed on the expressway.

The journey takes Nadia a total of 50 minutes.

Show that x² − 102x + 1440 = 0 . [5]
Solve the equation x² − 102x + 1440 = 0 .
Give your solutions correct to two decimal places. [3]
Explain why one of the solutions in part (b) must be rejected. [1]
Calculate the difference between the times Nadia spent driving on the expressway and driving on local roads.
Give your answer in minutes and seconds, correct to the nearest second. [2]

Worked solution

(a) The local-road speed is (x − 30) km/h, and 50 minutes is $\frac{50}{60} = \frac{5}{6}$ of an hour.

$\frac{40}{x} + \frac{20}{x - 30} = \frac{5}{6}$

Multiplying by 6x(x − 30):

240(x − 30) + 120x = 5x(x − 30)

240x − 7200 + 120x = 5x² − 150x

5x² − 510x + 7200 = 0, and dividing by 5,

x² − 102x + 1440 = 0 (shown)

(b) b² − 4ac = (−102)² − 4(1)(1440) = 10 404 − 5760 = 4644

$x = \frac{102 \pm \sqrt{4644}}{2} = \frac{102 \pm 68.146900…}{2}$

x = 85.07 or x = 16.93 (2 d.p.)

(c) If x = 16.93 then the local-road speed x − 30 is −13.07 km/h. A speed cannot be negative, so x = 16.93 must be rejected.

(d) Using the unrounded x = 85.073450…:

Time on the expressway $= \frac{40}{85.073450…}$ = 0.470181… h = 28.210916… min

Time on local roads $= \frac{20}{55.073450…}$ = 0.363151… h = 21.789083… min

Difference = 6.421833… min = 6 minutes and 25.309… seconds

6 minutes 25 seconds (nearest second)

Check: 28.210916… + 21.789083… = 50 minutes exactly, as the stem says.

Why this works. Five marks for (a): the setting up is the question. "30 km/h slower" becomes the denominator (x − 30), and 50 minutes becomes $\frac{5}{6}$ of an hour, since the speeds are in km/h. In (c), 16.93 is a reasonable speed; it goes because it makes the second speed negative, and that sentence is the mark, while (d) wants a difference, computed from the unrounded 85.073450…

The whole chapter in one table

If you can reproduce this table from memory, you have the skeleton of the topic. Everything else hangs off it.

	What you are given	What you write	Where the marks go
Factorisation ⓘ Recap	`ax`² + `bx` + `c` = 0 that factorises	(`x` − `r`)(`x` − `s`) = 0, then each bracket = 0	the factorised line, and both exact roots
Completing the square	`x`² + `px` + `q` (coefficient of `x`² = 1)	add (`p`/2)² to both sides → (`x` + `p`/2)² = … → take ±√ of both sides	the perfect-square line; keeping the ±
Quadratic formula	any quadratic, especially `a` ≠ 1	`a`, `b`, `c` → substitute → simplify the surd → round	the substitution line, with negatives in brackets
Graphical method	a table of values, or a drawn graph	plot, smooth curve, read the `x`-intercepts	the smooth curve; answers to 1 d.p.
Fractional equations	an equation containing algebraic fractions	× LCM of the denominators → expand → = 0 → solve	multiplying every term; rejecting a root that makes a denominator zero
Sketching: factorised form ✕ Not in G2	`y` = ±(`x` − `a`)(`x` − `b`)	`x`-intercepts `a` and `b`; turning point at `x` = (`a` + `b`)/2	all four critical points labelled as coordinates
Sketching: completed-square form ✕ Not in G2	`y` = ±(`x` − `p`)² + `q`	turning point (`p`, `q`); line of symmetry `x` = `p`	the turning point read off correctly, signs included
Word problems	a real-world situation	define `x` → everything in terms of `x` → form the equation → solve → reject → answer with units	the formulation, and the reason for the rejection

The formulae to know by heart

Completing the square. To complete the square for x² + bx, add

{(\frac{b}{2})}^{2}

, so that:

x^{2} + b x = {(x + \frac{b}{2})}^{2} - {(\frac{b}{2})}^{2}

The quadratic formula. The solutions of ax² + bx + c = 0, where a ≠ 0, are:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

The discriminant b² − 4ac. Positive → two real solutions, the curve cutting the x-axis twice. Zero → one repeated solution, the curve touching the axis. Negative → no real solutions, the curve missing the axis.

Graphs of y = ±(x − a)(x − b). ✕ Not in G2 The x-intercepts are a and b, and the line of symmetry

x = \frac{a + b}{2}

is also the turning point's x-coordinate.

Graphs of y = ±(x − p)² + q. ✕ Not in G2 The turning point is (p, q), with line of symmetry x = p. A leading + gives a minimum; a leading −, a maximum.

The habits that earn the marks

1. Show every line — "omission of essential working will result in loss of marks".
2. Round once, at the end: 3 s.f. unless the question names another accuracy.
3. Give both roots, unless there is a reason to reject one — and state the reason.
4. Rearrange to = 0 before applying any method.
5. In a fractional equation, check no answer makes a denominator zero.
6. Answer the question that was asked, in context, with units.

Syllabus at a glance

The fourthree G3 Mathematics (K310) outcomes this chapter covers, quoted from the Secondary Three/Four section of the syllabus.

G3 Mathematics (K310, first examination 2027) — these notes are written to it.

LO	What you must be able to do	Covered in
7.11	solving quadratic equations in one variable by: use of formula completing the square for `y` = `x`² + `px` + `q` graphical method	2.1 Completing the square, 2.1 The formula, 2.3 Graphical method
7.12	solving fractional equations that can be reduced to quadratic equations such as $\frac{6}{x + 4} = x + 3$ ; $\frac{1}{x - 2} + \frac{2}{x - 3} = 5$	2.2 Fractional equations
7.14	formulating a quadratic equation in one variable to solve problems	2.4 Problems & modelling
6.8	sketching the graphs of quadratic functions given in the form: `y` = (`x` − `p`)² + `q` `y` = −(`x` − `p`)² + `q` `y` = (`x` − `a`)(`x` − `b`) `y` = −(`x` − `a`)(`x` − `b`) ✕ Not in G2	2.3 Sketching from a given form

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus. Factorisation is not listed under 7.11 for G3 — it is assumed from Secondary 1/2 — but it is still the fastest route whenever it works, so it is recapped in 2.1. The G2 syllabus wording differs; a G2-specific outcome list for this chapter is authored in a later phase, and the parts G2 does not cover are already marked ✕ Not in G2.

📐 Quadratic Equations & Graphs — Revision Notes

The four methods — and how the paper tells you which one

Factorisation — the recap ⓘ Recap

Past-paper questions

Completing the square

Past-paper questions

The quadratic formula

Check yourself

Fractional equations that reduce to quadratics

Check yourself

Past-paper questions

What a quadratic graph looks like

Drawing a graph is not the same as sketching one

Past-paper questions

Sketching from a given form ✕ Not in G2

Solving a quadratic equation graphically

Check yourself

Formulating a quadratic from a word problem

Check yourself

Past-paper questions

The whole chapter in one table

The formulae to know by heart

The habits that earn the marks

Syllabus at a glance

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