中文（繁體）
中文（简体）
English
HKDSE Maths | Quadratic equations and the discriminant teaching guide | delta = 16, roots 1 and 5, k = plus or minus 6
Students can use the formula, then freeze when a question asks for the range of k. The fix is to treat the discriminant as a switch. Two numbers, delta = 16 and k = plus or minus 6, make the three cases usable.
Objectives — Students can: (1) solve x squared minus 6x plus 5 equals 0 by factorising, completing the square and the formula, and say when each is best; (2) use delta equals b squared minus 4ac to classify the roots and to work backwards for k; (3) check answers with the sum and product of roots and model a perimeter-and-area problem.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Factorising | How does x squared minus 6x plus 5 factorise? | (x – 1)(x – 5) |
| Perfect squares | What is (x – 3) squared expanded? | x squared minus 6x plus 9 |
| Square roots of negatives | Is the root of minus 4 real? | No, so there is no real root |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-10 min | Three methods | Factorising, completing the square, formula | Do the same equation three ways |
| 10-22 min | The discriminant | Delta equals b squared minus 4ac, three cases | Delta 16 gives two distinct roots |
| 22-32 min | Finding k | Set delta to zero and solve | k squared 36 gives plus or minus 6 |
| 32-42 min | Sum and product | Minus b over a, c over a | Check with 6 and 5 |
| 42-54 min | Modelling | Define a variable, build the equation, reject invalid roots | Perimeter 20, area 21 gives 3 and 7 |
| 54-60 min | Exit check | Three questions in class | Delta, k and roots all correct |
03 Teaching script (read it out)
Scene 1: Act 1: three methods, one equation, three times
Write x squared minus 6x plus 5 equals 0. Factorising gives (x – 1)(x – 5) equals 0, so 1 and 5.
Completing the square: (x – 3) squared minus 4 equals 0, so (x – 3) squared equals 4, x – 3 equals plus or minus 2, again 1 and 5. Point out the vertex at (3, -4) and the minimum value -4.
Formula: delta equals 36 minus 20 equals 16, root 16 is 4, x equals (6 plus or minus 4) over 2, giving 5 and 1.
The lesson is not which is fastest but that factorising is quick when visible, completing the square reveals the graph, and the formula always works.
Completing the square: (x – 3) squared minus 4 equals 0, so (x – 3) squared equals 4, x – 3 equals plus or minus 2, again 1 and 5. Point out the vertex at (3, -4) and the minimum value -4.
Formula: delta equals 36 minus 20 equals 16, root 16 is 4, x equals (6 plus or minus 4) over 2, giving 5 and 1.
The lesson is not which is fastest but that factorising is quick when visible, completing the square reveals the graph, and the formula always works.
All three methods must agree, or redo the work.
Scene 2: Act 2: the discriminant is a switch
Delta greater than zero: two distinct real roots, the curve crosses the x-axis twice. Delta equals zero: one repeated root, the curve touches the axis. Delta less than zero: no real roots.
Reverse question: x squared plus kx plus 9 equals 0 has equal roots. Set k squared minus 36 equals 0, so k equals 6 or -6. Insist on both; most students write only 6.
Then a range question: two distinct real roots requires k squared minus 36 greater than zero, so k greater than 6 or k less than -6. Draw the number line to show why it is “or”.
Reverse question: x squared plus kx plus 9 equals 0 has equal roots. Set k squared minus 36 equals 0, so k equals 6 or -6. Insist on both; most students write only 6.
Then a range question: two distinct real roots requires k squared minus 36 greater than zero, so k greater than 6 or k less than -6. Draw the number line to show why it is “or”.
The discriminant is a switch: positive, zero, negative.
Scene 3: Act 3: sum and product, then a real problem
The sum of roots is minus b over a, which is 6, and the product is c over a, which is 5. A two-second check catches most sign errors.
Modelling: a rectangle has perimeter 20 and area 21. Let the length be x, so the width is 10 minus x, and x(10 minus x) equals 21, giving x squared minus 10x plus 21 equals 0. Delta equals 16, so x equals (10 plus or minus 4) over 2, giving 7 or 3, which are the two sides.
Remind them to state units, check that lengths are positive, and answer the question that was asked.
Modelling: a rectangle has perimeter 20 and area 21. Let the length be x, so the width is 10 minus x, and x(10 minus x) equals 21, giving x squared minus 10x plus 21 equals 0. Delta equals 16, so x equals (10 plus or minus 4) over 2, giving 7 or 3, which are the two sides.
Remind them to state units, check that lengths are positive, and answer the question that was asked.
State units and reject the negative root.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Solving | Use the formula | Write a, b, c, compute delta, substitute | x equals (6 plus or minus 4) over 2 |
| Classifying | Describe the roots | Compute delta and compare with zero | Delta 16 means two distinct real roots |
| Unknown coefficient | Find k for equal roots | Set delta to zero and keep both signs | k equals 6 or -6 |
| Modelling | Turn perimeter and area into an equation | One variable, the other from the total | x squared minus 10x plus 21 equals 0 |
Sanity check (teach this): Most delta errors are sign errors: with b equal to -6, b squared is 36, not -36.
Delta equals zero means one repeated root, not no root.
Every square root carries plus or minus, so k squared equals 36 gives two values.
Return to the context: a length cannot be negative, so check both roots.
Delta equals zero means one repeated root, not no root.
Every square root carries plus or minus, so k squared equals 36 gives two values.
Return to the context: a length cannot be negative, so check both roots.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Roots of x squared minus 6x plus 5 | 1 and 5 | (x – 1)(x – 5) equals 0 |
| Value of delta | 16 | 36 minus 20 |
| k when x squared plus kx plus 9 has equal roots | 6 or -6 | k squared minus 36 equals 0 |
| Sides of the rectangle, perimeter 20 and area 21 | 7 and 3 | x squared minus 10x plus 21 equals 0 |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Dropping the plus or minus | Answering only k equals 6 | “A square has two roots: k squared 36 means two answers, not one.” |
| Calling delta zero no solution | Answering that there are no roots | “Delta zero is one repeated root; the curve just touches the axis.” |
| Sign slip on b squared | Writing delta as -16 | “b is negative, so its square must be positive.” |
| Keeping a negative length | Adding -3 to the sides | “A negative length does not exist, so reject it even if the algebra gives it.” |
07 Homework (paper, section and question numbers)
- Secondary maths chapter index — Questions 1-5, three methods
- All secondary prep guides — Questions 6-8, discriminant and k
- DSE maths practice — Question 9, modelling with units
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Value of delta | 16 | Wrong means recompute b squared |
| k for equal roots | plus or minus 6 | Missing one sign means reteach square roots |
| Sides of the rectangle | 7 and 3 | Wrong means repeat the modelling steps |
09 Tutor FAQ
Do students need all three methods?
Yes. Factorising is fastest, completing the square shows the graph, and the formula always works.
Does delta less than zero mean no solution?
No real solution; complex roots exist, but DSE only asks for real roots.
Why is the sum of roots minus b over a?
Because (x – alpha)(x – beta) expands to x squared minus the sum times x plus the product; compare coefficients.
Find a tutor — freeTutor Registration →
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.
