A quadratic equation is not difficult because it contains a squared letter. It becomes difficult when the student does not know which form will reveal the solutions. Sometimes two brackets make the answer visible. Sometimes a completed square explains both the answer and the graph. Sometimes the quadratic formula is the cleanest route. In a word problem, none of those methods can help until the situation has been translated correctly.
This Secondary 3 Mathematics Learning Guide teaches the whole journey: recognising a quadratic relationship, selecting a method, solving accurately, deciding which solutions are allowed, and checking the result against the original question. It follows the broader Algebraic Control guide and belongs to the Secondary Mathematics Hub.
Scope: the official 2027 SEC G3 Mathematics syllabus, section N7, includes quadratic solving and equation formulation. This is a G3-oriented topic guide, not a timetable for every Secondary 3 class. Follow the methods and scope assigned by your school.
Choose your starting point: diagnostic · method choice · word problems · practice · explained answers. The examples and exercises below are original teaching questions, not reproduced examination questions.
What Makes an Equation Quadratic?
A quadratic equation in one unknown can be written as ax² + bx + c = 0, with a ≠ 0. The letters a, b and c represent fixed coefficients; x is the unknown. The condition a ≠ 0 matters: without a squared term, the equation would no longer be quadratic. A coefficient may be negative, fractional or zero, except that a cannot be zero.
The equation x² − 9 = 0 is quadratic even though no x term is written. Its coefficients are a = 1, b = 0 and c = −9. Similarly, 3x² = 12x is quadratic, but it is not yet in the standard form needed for a straightforward method comparison. Subtracting 12x gives 3x² − 12x = 0.
Be careful with appearance. In (x + 2)² = x² + 9, expanding and cancelling the squared terms gives 4x + 4 = 9. The resulting equation is linear, not quadratic. Classification should follow valid simplification rather than the first symbol you notice.
Expression, Equation and Function: Three Different Jobs
The expression x² − 5x + 6 can be factorised or evaluated, but it does not ask for a solution on its own. The equation x² − 5x + 6 = 0 asks which inputs make that expression zero. The function y = x² − 5x + 6 describes an output for each input. The algebra looks similar because the same expression is involved, but the requested job changes.
For the equation, x = 2 and x = 3 are solutions. For the function, (2, 0) and (3, 0) are points on the graph. The corresponding x-values locate its x-intercepts. Saying that the answers are “(2, 0) and (3, 0)” when asked to solve for x confuses coordinate answers with numerical solutions.
A Five-Question Diagnostic Before the Main Lesson
Try these without looking ahead. Expand (x − 3)(x + 2). Solve (x − 3)(x + 2) = 0. Solve x² = 16. Write x² + 6x as a square plus or minus a constant. Finally, a rectangle has width w cm and length w + 3 cm: write an equation saying its area is 40 cm².
The answers are x² − x − 6; x = 3 or −2; x = 4 or −4; (x + 3)² − 9; and w(w + 3) = 40. These questions check different dependencies. A wrong expansion points to bracket or sign work. A missing negative solution points to square-root reasoning. Difficulty forming the rectangle equation points to modelling, even when algebraic procedures are fluent.
Do not interpret one error as evidence that the entire topic must be restarted. Repair the particular dependency, then return to the problem that needed it. For example, expand and check three products before attempting factorisation again; do not spend an hour repeating correct linear-equation work.
How to Choose a Quadratic Method
| What you see | Useful first move |
|---|---|
| A common factor in every term | Extract it before doing more work. |
| An isolated square | Use square roots, remembering both signs. |
| A readily factorisable expression equal to zero | Factorise and use the zero-product property. |
| x² + px + q, especially when a turning point matters | Consider completing the square. |
| No convenient factorisation | Use the quadratic formula. |
| A graph with an instruction to estimate | Read the required intersections to suitable precision. |
This is a choice guide, not a competition to avoid the formula. The best method is the one that fits the question and that you can execute reliably. When a question specifies a method, use that method. When it does not, inspect briefly rather than spending several minutes searching for integer factors that may not exist.
Why Factorisation Works Only After the Product Equals Zero
If two real numbers multiply to zero, at least one must be zero. To see why, suppose the first number is not zero. Dividing the product equation by that nonzero number forces the second number to be zero. This is the zero-product property, and it is the reason factorisation solves equations.
It does not say that every product lets us set individual factors equal to the right-hand side. From (x − 2)(x + 5) = 12, you cannot conclude that x − 2 = 12 or x + 5 = 12. First expand, bring all terms to one side, and obtain an equation whose right-hand side is zero. The special role of zero is the mathematical justification, not a formatting preference.
Worked Example 1: Factorisation With Two Distinct Roots
Solve x² − 5x + 6 = 0. We need two numbers with product 6 and sum −5. Both must be negative, so use −2 and −3. Then (x − 2)(x − 3) = 0. Therefore x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.
Check each answer in the original equation. At x = 2, 4 − 10 + 6 = 0. At x = 3, 9 − 15 + 6 = 0. Checking only one root leaves the other unverified. The word “or” is useful because either value separately satisfies the equation; x is not being assigned two values simultaneously in one substitution.
Worked Example 2: The Coefficient of x² Is Not One
Solve 6x² + x − 2 = 0. One systematic approach is to split the middle term. The product of the first and last coefficients is −12. Two numbers with product −12 and sum 1 are 4 and −3. Thus 6x² + 4x − 3x − 2 = 0.
Group the terms: 2x(3x + 2) − 1(3x + 2) = 0. The common bracket produces (2x − 1)(3x + 2) = 0. Hence x = 1/2 or x = −2/3. Re-expansion gives 6x² + 4x − 3x − 2, so both the middle coefficient and constant have been checked.
A common error is reporting 1 and −2 because those numbers appear inside the brackets. Each bracket still forms a small linear equation. Solve 2x − 1 = 0 and 3x + 2 = 0 rather than reading off constants without their coefficients.
Worked Example 3: Do Not Divide Away a Solution
Solve 3x² = 12x. Dividing both sides by x immediately gives 3x = 12 and x = 4, but that step assumes x ≠ 0. Nothing in the question excludes zero. Indeed, zero satisfies the original equation.
The reliable route is 3x² − 12x = 0, then 3x(x − 4) = 0. Therefore x = 0 or x = 4. Dividing by a variable is not automatically wrong, but it requires a separate check of the case where that variable is zero. Factorisation keeps both cases visible without that extra branching.
Worked Example 4: Completing the Square From Meaning
Solve x² + 6x − 7 = 0 by completing the square. The expansion (x + 3)² = x² + 6x + 9 tells us exactly what is missing. Replacing x² + 6x with (x + 3)² − 9 preserves the expression. Consequently, the equation becomes (x + 3)² − 16 = 0.
Now (x + 3)² = 16, so x + 3 = 4 or x + 3 = −4. Subtracting 3 gives x = 1 or x = −7. Check: 1 + 6 − 7 = 0, and 49 − 42 − 7 = 0.
The instruction “halve the coefficient and square it” is a shortcut to the expansion, not an unexplained ritual. If the coefficient of x is 6, the bracket must contain 3 because the cross term in (x + 3)² is 2 × 3 × x. Adding 9 without compensating changes the expression; completing the square must preserve equality throughout.
Completed-Square Form Also Explains the Graph
The function y = x² + 6x − 7 can also be written y = (x + 3)² − 16. Since a real square is never negative, the smallest output is −16, occurring when x + 3 = 0. The minimum point is therefore (−3, −16), and the line of symmetry is x = −3.
The roots 1 and −7 lie equally far from −3: each is four units away. This gives an independent structural check on the solutions. It also illustrates why solving and graph interpretation belong together. The equation gives the inputs producing zero; the graph shows where those inputs sit relative to the turning point. A graph estimate is useful, but it should not replace an exact result when exact values are requested.
Worked Example 5: The Quadratic Formula Without Sign Errors
For ax² + bx + c = 0, the quadratic formula is x = [−b ± √(b² − 4ac)]/(2a). The entire numerator is divided by 2a. Brackets are especially important when using a calculator or when b and c are negative.
Solve 2x² − 3x − 4 = 0. Here a = 2, b = −3 and c = −4. The expression under the square root is (−3)² − 4(2)(−4) = 9 + 32 = 41. Therefore x = (3 + √41)/4 or x = (3 − √41)/4.
These are exact answers. To three significant figures, they are approximately 2.35 and −0.851. Keep the exact expressions or calculator precision for checking; substituting a rounded decimal will usually produce a small nonzero residual rather than exactly zero. That residual alone does not mean the exact root was wrong.
Before calculating, predict the signs. The product of the roots is c/a = −2, so the two real roots must have opposite signs. This is a useful checking extension derived by expanding a(x − r)(x − s): its constant is ars. It is not necessary to memorise extra rules before understanding the main formula.
One Root, Two Roots or No Real Roots?
The quantity b² − 4ac determines whether the square-root part of the formula is positive, zero or negative. A positive value gives two distinct real roots. Zero makes the plus and minus cases coincide, giving one repeated real root. A negative value gives no real roots. This is an explanation of the formula’s behaviour; use any additional terminology according to your course.
For (x − 3)² = 0, the only solution is x = 3. Repetition means that the same factor occurs twice, not that there are two different answers. For (x + 1)² + 4 = 0, the left side is at least 4 for every real x, so no real solution exists. A calculator error message is not an explanation; the non-negativity of a square is.
Word Problems: Build the Relationship Before Solving It
A useful modelling sequence is to name the unknown with units, express dependent quantities using that unknown, state the condition connecting them, solve the resulting equation, and interpret the roots. The equation should be readable back into the story. “x(x + 3) = 40” should mean “width multiplied by length equals area”, not merely “these numbers seemed to fit”.
The domain is the set of allowed inputs. A length must be positive. A count may have to be a whole number. A time measured after the start cannot be negative. These restrictions decide which algebraic solutions answer the actual question, but they must come from the question rather than an assumption that negative roots are always wrong.
Worked Example 6: Rectangle Dimensions and a Rejected Root
A rectangle has length x + 4 cm and width x − 1 cm. Its area is 66 cm². Find its dimensions. Both dimensions must be positive, so x > 1. The area relationship gives (x + 4)(x − 1) = 66. Expand to get x² + 3x − 4 = 66, hence x² + 3x − 70 = 0.
Factorising gives (x + 10)(x − 7) = 0. Thus x = −10 or x = 7. Reject −10 because it violates x > 1. With x = 7, the length is 11 cm and the width is 6 cm. Check the requested dimensions, not just x: 11 × 6 = 66 cm², and their difference is 5 cm as the original expressions imply.
The answer “7 cm” would be incomplete. The variable was an intermediate description, not the length or width requested. Always reread the final sentence after solving.
Worked Example 7: A Border Adds Width on Both Sides
A 12 m by 8 m rectangular garden is surrounded externally by a uniform path. The combined area of garden and path is 192 m². Find the path’s width. Let the width be x m, with x > 0. The outer dimensions are 12 + 2x and 8 + 2x, because the path appears on both sides of each dimension.
Thus (12 + 2x)(8 + 2x) = 192. Expanding gives 96 + 40x + 4x² = 192. Divide the rearranged equation by 4 to obtain x² + 10x − 24 = 0. Factorising gives (x + 12)(x − 2) = 0. The positive solution is x = 2 m.
The outer dimensions are then 16 m and 12 m, whose product is 192 m². The path itself occupies 192 − 96 = 96 m². Notice the distinction between “combined area” and “path area”: using the wrong one would create the wrong equation even with flawless algebra.
Worked Example 8: Speed and Time Produce a Quadratic
A cyclist completes a 60 km route. Increasing the constant speed by 5 km/h would reduce the journey time by one hour. Find the original speed. This is an idealised constant-speed model. Let the original speed be v km/h, with v > 0. The original time is 60/v hours, and the faster time is 60/(v + 5) hours.
The difference in times is 60/v − 60/(v + 5) = 1. Multiplying by v(v + 5), which is nonzero for the allowed speeds, gives 60(v + 5) − 60v = v(v + 5). Consequently, 300 = v² + 5v, so (v + 20)(v − 15) = 0.
The physically allowed speed is 15 km/h. At 15 km/h the journey takes four hours; at 20 km/h it takes three hours. Their difference is exactly one hour. The negative algebraic root is rejected because speed in this model must be positive, not because negative answers are generally unacceptable in Mathematics.
For more work on clearing denominators without losing restrictions, use the companion guide on Algebraic Fractions and Formula Rearrangement.
Four Errors That Look Similar but Need Different Repairs
Wrong equation: the student writes 12 + x instead of 12 + 2x for the border. Repair the diagram and the meaning of the width. More factorisation will not fix this modelling error.
Wrong factorisation: the expanded coefficients do not match the original. Repair product-and-sum reasoning and re-expansion. Do not teach the student to accept brackets merely because they look plausible.
Missing root: the student divides by x or forgets the negative square root. Repair the validity of the transformation and the full set of solutions. This is different from an arithmetic slip.
Wrong interpretation: the algebra is correct, but a negative width is accepted or only x is reported when dimensions were asked for. Repair the return to context. A complete solution must satisfy both the equation and the original task.
Independent Practice: Choose the Method Yourself
For Questions 1–8, solve over the real numbers. Write a method choice before your first calculation, and check every proposed solution. For Questions 9–12, define the unknown and state its restrictions. No marks or grade boundaries are implied by this practice set.
1. x² − 9 = 0.
2. x² + 2x − 15 = 0.
3. 2x² − 7x + 3 = 0.
4. x² − 8x + 7 = 0, using completing the square.
5. 3x² − 2x − 5 = 0.
6. x² + 4x + 1 = 0; leave exact answers.
7. 4x² − 12x + 9 = 0.
8. x² + 2x + 5 = 0.
9. A rectangle has width w cm and length w + 3 cm. Its area is 40 cm². Find both dimensions.
10. The product of two consecutive positive integers is 156. Find the integers.
11. A model gives a ball’s height as h = −5t² + 20t + 1 metres, where t is time in seconds and 0 ≤ t ≤ 4. At what times is the height 16 m?
12. A student solves x² = 5x by dividing by x and reports x = 5. Explain the missing case and give the full solution.
Explained Answers to the Practice Set
1. x² = 9 gives x = 3 or −3. Both square to 9. A positive square does not imply a positive input.
2. (x + 5)(x − 3) = 0, so x = −5 or 3. The factor constants must multiply to −15 and add to 2.
3. (2x − 1)(x − 3) = 0, so x = 1/2 or 3. Substituting 1/2 gives 1/2 − 7/2 + 3 = 0.
4. x² − 8x + 7 = (x − 4)² − 9. Therefore (x − 4)² = 9 and x = 4 ± 3, giving 1 or 7. This also places the roots symmetrically around x = 4.
5. (3x − 5)(x + 1) = 0. Hence x = 5/3 or −1. Expanding the product gives 3x² + 3x − 5x − 5, which confirms the middle term.
6. Completing the square gives (x + 2)² − 3 = 0. Thus x = −2 ± √3. These exact values should not be replaced by decimals when exact answers are requested.
7. (2x − 3)² = 0, so x = 3/2 is the only distinct solution. The equation has a repeated root.
8. (x + 1)² + 4 = 0 has no real solution because its left-hand side is always at least 4. “No real solution” is the conclusion with its reason, not an unfinished calculation.
9. w(w + 3) = 40 gives (w + 8)(w − 5) = 0. Since w > 0, take w = 5. The rectangle is 5 cm by 8 cm, and 5 × 8 = 40.
10. Let the smaller integer be n. Then n(n + 1) = 156, so (n + 13)(n − 12) = 0. The positive integer solution is n = 12; the integers are 12 and 13.
11. Set the height equal to 16: −5t² + 20t + 1 = 16. Rearranging and dividing by −5 gives t² − 4t + 3 = 0. Therefore t = 1 or 3 seconds. Both lie inside the stated interval, and both give height 16 m. A contextual question can legitimately retain both roots.
12. Dividing by x excluded x = 0 without checking it. Instead write x(x − 5) = 0. The complete answer is x = 0 or x = 5, and both satisfy the original equation.
From Corrected Work to Independent Work
After correcting a question, close the solution and explain why the chosen method works. On another occasion, attempt a changed version without the chapter heading. Change a coefficient, replace an area statement with a time statement, or ask for dimensions instead of x. The purpose of this suggested routine is to test whether the mathematical decision survives a changed surface.
A compact error record needs only the question, the first invalid step, the reason it was invalid, and one check that would have detected it. “Careless” is not an actionable diagnosis. “I divided by a quantity that could be zero” is actionable because it identifies a precise decision to change next time.
Teacher and Parent Prompts That Preserve the Thinking
When a learner is stuck, start with “What does x represent?” or “Which condition gives the equation?” before supplying a formula. If the equation is already correct, ask “What form would make the solutions visible?” If a root is rejected, ask “Which sentence of the question excludes it?” These prompts distinguish modelling, technique and interpretation rather than treating all hesitation as the same difficulty.
For a learner who needs support, use one readily factorisable example and require complete explanation. For a learner ready for extension, ask for the same equation to be solved in two ways, or ask for an equation with specified roots. Roots 2 and −5, for example, lead to (x − 2)(x + 5) = 0 and hence x² + 3x − 10 = 0. Both learners are studying the same underlying structure at different levels of independence.
Questions Students Often Ask
Must a quadratic equation always have two answers? It has at most two distinct real roots. A repeated root gives one distinct real answer, while some quadratics have no real roots.
Can I always use the formula? It applies to a genuine quadratic with a ≠ 0, but a question may require a particular method. Even without a restriction, an isolated square or obvious factorisation may be clearer and quicker.
Should I always reject a negative answer? No. Reject an answer only when it violates the domain or another condition. Negative coordinates and negative number solutions can be valid; negative physical lengths are not.
Is a correct calculator answer enough? A numerical output does not show whether the original relationship was represented correctly. Record the equation, the method and the interpretation so the answer can be checked.
Continue the Secondary 3 Learning Route
Use Algebraic Fractions and Formula Rearrangement for restrictions and fractional equations; Indices, Standard Form and Estimation for powers and numerical control; and Similarity, Scale Factors and Mensuration for modelling relationships involving length, area and volume.
The complete quadratic solution is more than a pair of numbers: it is a justified equation, a suitable method, every valid root, and an answer that still belongs to the original situation. Return to the Secondary Mathematics Hub to choose the next capability to strengthen.
