S3 Maths — Quadratic Equations in Context: Speed, Geometry and Rejecting a Root (Lesson Prep Guide)

2026-11-15

Tutor lesson-prep guides · Secondary Maths

Students can solve x squared plus 3x minus 40 equals zero, then freeze at a rectangle whose length is 3 cm more than its width with area 40 square centimetres, because they cannot separate the unknown from the condition. This lesson does three things only: name the unknown, build the equation, reject the root that cannot be real.

Objectives — Students can: (1) choose one unknown from a word problem and express the remaining data in terms of it; (2) build a quadratic equation from either an area model or a time-equals-distance-over-speed model and solve it; (3) explain why one root must be rejected and verify the remaining root by substituting it back into the original wording.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Solving quadratics	What are the roots of x squared minus 5x plus 6 equals zero?	2 and 3
Factorising	How does x squared plus 3x minus 40 factorise?	(x + 8)(x – 5)
Speed formula	Write the relation between speed, time and distance	Time equals distance divided by speed

02 Lesson flow (45 minutes)

Time	Stage	Tutor does	Student does
0-5 min	A stuck moment	Write length is 3 more than width, area 40, and ask what the first step is	Name the width as x
5-14 min	Setting the unknown	Show the length as x + 3 and have students fill a table of width, length, area	Write the three expressions
14-29 min	Two models	Do the area model x(x + 3) = 40, then the speed model 60/v + 60/(v + 10) = 5	Say which formula each model uses
29-38 min	Rejecting a root	Both roots appear, one negative; ask whether a width can be negative	Write the rejection sentence
38-42 min	Independent practice	Three questions on area, consecutive integers, right triangles; tour asking which quantity is x	Name the unknown before writing
42-45 min	Exit check	Collect answers and ask why the second root is dropped	It lies outside the possible values

03 Teaching script (read it out)

Scene 1: The first step is always the unknown

Do not copy numbers straight from the question. Ask what the question wants you to find, call that x, and express everything else in terms of x. Length is 3 more than width means length equals x + 3, not length equals 3. Area 40 means x times (x + 3) equals 40, and that is the equation. Ask each student to say in one sentence what they called x.

Students state which quantity they called x

Scene 2: Two models: area and time

The area model is always length times width equals area. The speed model has one route only: time equals distance divided by speed, written separately for each leg and then added. A 60 km outward trip at speed v takes 60 divided by v; the return trip at v + 10 takes 60 divided by (v + 10). The two times add to 5 hours, and the equation is ready. The common slip is turning 10 km per hour faster into 10 hours less, so make them read the units aloud.

Students write the two times separately

Scene 3: Only one root fits the real world

A quadratic always gives two roots, but lengths, speeds and times cannot be negative. From x = 5 and x = -8 students must write: x is a width, so it cannot be negative, therefore x = 5. That sentence is a mark, not politeness. Finish by substituting back: width 5, length 8, area 40, which matches the question.

Students explain why the negative root is rejected

04 Question types and worked steps

Type	Example	Steps	Answer
Area model	A rectangle has length 3 cm more than its width and area 40 square centimetres. Find the width.	1) Width x, length x + 3. 2) Write x(x + 3) = 40. 3) Rearrange to x squared plus 3x minus 40 equals zero and factorise	Width 5 cm, length 8 cm
Consecutive integers	The product of two consecutive positive integers is 72. Find the smaller one.	1) Smaller x, larger x + 1. 2) Write x(x + 1) = 72. 3) Solve x squared plus x minus 72 equals zero	x = 8; the root minus 9 is rejected
Speed and time	A car covers 60 km at speed v and returns 10 km per hour faster, taking 5 hours in total. Find v.	1) Outward time 60 over v. 2) Return time 60 over (v + 10). 3) Add to 5 and multiply by v(v + 10)	v = 20 km per hour; minus 6 is rejected
Right triangle	The two legs of a right triangle differ by 7 cm and the hypotenuse is 13 cm. Find the shorter leg.	1) Shorter leg x, longer x + 7. 2) Use x squared plus (x + 7) squared equals 169. 3) Simplify to x squared plus 7x minus 60 equals zero	5 cm; the root minus 12 is rejected

Sanity check (teach this): Three checks: first, substitute the answer into every sentence of the question, so the difference and the area must both match; second, count the roots, because a quadratic always has two and seeing one usually means a term was lost; third, test the range, since length, speed and time must be positive and a negative root needs a written rejection. Numeric anchors: 5 and 8 differ by 3 and multiply to 40, and 5, 12 and 13 satisfy 25 plus 144 equals 169, so either pair can be used to check a student answer on the spot.

05 Independent practice (3 questions)

Question	Answer	What to watch
The product of two consecutive positive integers is 72. Find the smaller one.	8	Check they used x and x + 1, not x and x + 2
A rectangle has length 3 cm more than its width and area 40 square centimetres. Find the width.	5 cm	Check they did not write the length as x minus 3
The legs of a right triangle differ by 7 cm and the hypotenuse is 13 cm. Find the shorter leg.	5 cm	Check they squared both legs

06 Common mistakes and what to say

Mistake	Student writes	Say this
Turning 3 more into minus 3	length equals x minus 3, so x(x – 3) = 40	The length is 3 more, so it is x plus 3, never minus
Faster speed means less time, minus 10	outward 60 over v, return 60 over v minus 10	The extra speed goes into the denominator: 60 over (v + 10)
Forgetting the rejection sentence	x = 5 or x = -8, answer 5 or -8	Write: width cannot be negative, so x = 5
Not substituting back	Stop at 5 and move on	Substitute back: length 8, 5 times 8 equals 40, which matches

07 Homework (paper, section and question numbers)

Quadratic formula and discriminant, question 3 — A positive discriminant is needed before two real roots exist
Solving simultaneous equations by elimination, question 2 — Same habit of naming the unknown before building equations
Polynomial operations, question 4 — A lost term makes the equation wrong from the start

08 Exit check (last 5 minutes)

Question	Answer	Decision
A number squared equals 6 times the number. Find the non-zero number.	6	Answering 0 means the word non-zero was missed
A speed equation gives v = 20 or v = -6. How do you write the answer?	20 km per hour, because a negative speed is impossible	Giving only 20 without the reason loses the mark
A rectangle has width x, length x + 5 and area 24. Find the width.	3	Answering 8 suggests width and length are swapped

09 Tutor FAQ

Must S3 students memorise the formula?

Most context questions here factorise neatly, so the formula is a backup. Use it only when factorising fails, as the discriminant chapter already covers it.

Should speed questions use a table?

Yes. Two rows for outward and return, three columns for distance, speed and time. Fill in what is known and the equation appears.

What if a student insists on two answers?

Have them read the answer aloud as a negative width in centimetres. Hearing it usually settles the point faster than an explanation.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.