Secondary Mathematics: Quadratic Equations, Factorisation and Roots

Secondary Mathematics · Worked Repair Guide 18

A quadratic equation is not finished when the brackets have been found. Factorisation changes the form of an expression. Solving identifies the permitted values of the unknown. Checking asks whether those values satisfy the original equation and its context. Confusing these three jobs is one reason a familiar quadratic can still produce an unreliable answer.

This worked guide follows the complete route: identify the equation, put it into a useful form, choose a method, find every candidate, and test the original conditions. Along the way, a quadratic is read through its factors, completed square, graph and discriminant. These are connected descriptions, not unrelated chapters.

The shorter diagnostic explanation remains at Why Do Quadratic Equations Feel Like Several Different Topics at Once? This companion adds detailed calculations, contrasting cases, modelling and independent practice. All numerical situations are original teaching examples, not actual measurements or official examination questions.

Starting points: attempt the entry check, study factorisation, use completing the square to understand the general formula, then try twenty independent tasks. Extensions should be matched to the learner’s current course.

1. Distinguish an expression, an equation and a function

The expression x² − 7x + 12 has a value for each real x. The equation x² − 7x + 12 = 0 asks which values make that expression zero. The function y = x² − 7x + 12 describes a whole collection of input-output pairs. A root of the equation corresponds to an x-intercept of the function’s graph.

Before reading on, factorise x² − 7x + 12, solve x² − 7x + 12 = 0, and evaluate the expression at x = 2. The answers are (x − 3)(x − 4), x = 3 or 4, and 2. The same expression supports three different tasks.

Now solve x² = 25 and evaluate √25. The equation has solutions 5 and −5, whereas the principal square root √25 is 5. Finally, decide whether x² + 4 = 0 has a real solution. It does not, because every real square is non-negative. These small checks reveal whether the learner is tracking the question rather than applying a remembered pattern automatically.

2. Identify the actual quadratic after simplifying

A quadratic equation can be written ax² + bx + c = 0 with a ≠ 0. The coefficient a cannot be zero, because that would remove the quadratic term. The coefficients b and c may be zero. This standard form and the common solution methods are described in OpenStax’s section on quadratic equations.

Do not classify an equation only by the first squared term you see. In x² + 3x = x² + 15, subtracting x² from both sides leaves 3x = 15, a linear equation with solution x = 5. The squared terms cancel.

Conversely, x(x + 2) = 15 is quadratic even though it is not initially expanded. It becomes x² + 2x − 15 = 0. The required form emerges from lawful simplification. This first inspection can save considerable work and prevents putting the wrong coefficients into the quadratic formula.

3. Factorisation exposes a product that can be zero

To factorise x² − 7x + 12, seek two numbers whose product is 12 and whose sum is −7. They are −3 and −4, so the expression is (x − 3)(x − 4). Expanding checks both the middle coefficient and the constant.

The equation becomes (x − 3)(x − 4) = 0. For a product of real numbers to be zero, at least one factor must be zero. Therefore x − 3 = 0 or x − 4 = 0, giving x = 3 or x = 4. The word “or” matters: x need not equal both numbers at the same time.

The underlying zero-product property can be justified directly. If uv = 0 and u ≠ 0, divide by u to obtain v = 0. The remaining case is u = 0. There is no third possibility. This logical split is why a product form is valuable when the equation equals zero.

4. The right side must be zero before using the zero-product property

Consider (x − 2)(x + 1) = 10. It is not valid to set x − 2 = 10 or x + 1 = 10. Products can equal 10 through many pairs of factors. The zero-product property is special to zero.

Expand and rearrange: x² − x − 2 = 10, hence x² − x − 12 = 0. Factorise to (x − 4)(x + 3) = 0. The solutions are x = 4 and x = −3. Check them in the original product: 2 × 5 = 10 and (−5)(−2) = 10.

This is an example of allowing a neat expression to become temporarily less neat so that the correct structure can emerge. Keeping the original brackets at all costs would make the solution harder. The target is a justified equation, not a permanently tidy page.

5. A non-unit leading coefficient requires all products to agree

Solve 2x² − 7x − 4 = 0. The product of the leading coefficient and constant is −8. Two numbers with product −8 and sum −7 are −8 and 1. Split the middle term:

2x² − 8x + x − 4 = 0
2x(x − 4) + 1(x − 4) = 0
(2x + 1)(x − 4) = 0.

Thus x = −1/2 or x = 4. The factor 2x + 1 gives −1/2, not −1. At x = −1/2, the original expression is 1/2 + 7/2 − 4 = 0. At x = 4 it is 32 − 28 − 4 = 0.

The split-middle-term method is one route, not a compulsory ritual. Trial factorisation also works when the coefficients are manageable. Whatever route is used, expand the proposed factors before extracting roots; a correct-looking pair of brackets can contain a wrong middle coefficient.

6. Do not divide away a possible zero solution

The equation 3x² − 12x = 0 factorises as 3x(x − 4) = 0. Its solutions are x = 0 and x = 4. Dividing the original equation by x without considering x = 0 would leave 3x − 12 = 0 and lose a solution.

Dividing by the non-zero constant 3 is harmless; dividing by the variable x is a different decision. Factorisation itself has introduced no restriction. It simply rewrites the expression as an equal product for every real x.

A practical repair is to circle any proposed divisor containing the unknown. Ask whether it can be zero in the original equation. When it can, either avoid that division or separate the zero case explicitly. Checking only the solution x = 4 afterwards will not reveal the lost root; completeness depends on the transformation chain as well as the final substitutions.

7. A squared expression may already be ready to solve

For (3x − 2)² = 25, there is no need to expand. The expression 3x − 2 must be 5 or −5. Therefore 3x = 7 or −3, giving x = 7/3 or x = −1.

Check the unsquared expressions: at 7/3, 3x − 2 = 5; at −1, it equals −5. Both squares equal 25. Taking only the positive possibility would discard one valid solution.

Compare (3x − 2)² = 0. Now the only possibility is 3x − 2 = 0, so x = 2/3. Compare (3x − 2)² = −4 over the reals: there is no solution. The same left side produces two, one or no distinct real roots depending on the right side. Inspecting this structure is more economical than applying the formula to every example.

8. Completing the square creates a readable centre

Take x² − 8x + 12. Half of −8 is −4, and (x − 4)² expands to x² − 8x + 16. To preserve the original constant, subtract 4:

x² − 8x + 12 = (x − 4)² − 4.

Solving the equation equal to zero gives (x − 4)² = 4, then x − 4 = ±2. The roots are 2 and 6. The completed square also shows the graph’s minimum value −4 at x = 4, since the square cannot be negative.

The common mistake is to add the number needed to complete the square without compensating for it. That changes the expression. An identity must preserve value for every input, not merely make the shape attractive. Expanding the final form gives an immediate check of the compensation.

9. Completing the square also handles awkward roots

Solve x² + 4x − 7 = 0. Rewriting gives (x + 2)² − 11 = 0. Thus (x + 2)² = 11 and x = −2 ± √11. These are exact real roots even though they are not integers or simple fractions.

A symbolic check is easier than a rounded decimal check: substituting either root into (x + 2)² − 11 gives (±√11)² − 11 = 0. A decimal approximation can be added if requested, but should not replace the exact form prematurely.

For 2x² + 8x + 3 = 0, factor 2 from the quadratic and linear terms: 2[(x + 2)² − 4] + 3 = 0. This becomes 2(x + 2)² − 5 = 0, so x = −2 ± √(5/2). Multiplying the completed square by 2 must also multiply the compensating term. That outer factor is a frequent source of errors.

10. The quadratic formula is completing the square in general

Begin with ax² + bx + c = 0, a ≠ 0. Divide by a, move c/a to the other side, and add the square of half the x-coefficient. The result is:

[x + b/(2a)]² = (b² − 4ac)/(4a²).

Taking both square-root possibilities and rearranging gives the familiar formula x = [−b ± √(b² − 4ac)]/(2a), provided the square root is real for the problem being solved. When a is negative, the two signs still give the same two candidates; the ± accounts for the sign change associated with the denominator.

The formula is therefore not a competing branch of mathematics. It is a compressed general calculation. Its denominator divides the entire numerator, and b is the signed coefficient after the equation has been arranged with zero on one side. Recording a, b and c before substitution protects both of those details.

11. Use the signed coefficients, including zero coefficients

For 3x² − 5x − 2 = 0, the coefficients are a = 3, b = −5 and c = −2. The discriminant is (−5)² − 4(3)(−2) = 25 + 24 = 49. Thus x = [5 ± 7]/6, giving x = 2 or x = −1/3.

The first numerator uses −b = 5. Inside b², the complete negative coefficient is squared. Writing −5² instead of (−5)² would compute a different number. Likewise, forgetting that c is negative would turn the discriminant into 1 rather than 49.

For 4x² − 9 = 0, b = 0. For 2x² + 7x = 0, c = 0. Missing terms have zero coefficients; they do not cause later coefficients to shift into the wrong slot. The simpler factorisation or square-root route may be faster, but the formula should agree when used correctly.

12. The discriminant predicts the number of real roots

The discriminant D = b² − 4ac is the expression under the square root. If D > 0, the plus and minus choices give two distinct real roots. If D = 0, they coincide, producing one distinct repeated root. If D < 0, there are no real roots.

For x² − 6x + 9 = 0, D = 0 and the expression is (x − 3)². For x² − 6x + 10 = 0, D = −4 and the expression is (x − 3)² + 1, which is always positive for real x. The completed-square form and discriminant agree.

“No real roots” is not the same as “an invalid equation”. It states the result in the chosen number system. Complex-number solutions belong to a different extension. This guide keeps its equation-solving conclusions in the real domain unless stated otherwise.

13. Match each algebraic feature to its graphical meaning

For y = x² − 8x + 12, the factors (x − 2)(x − 6) give x-intercepts (2, 0) and (6, 0). The completed square (x − 4)² − 4 gives vertex (4, −4) and symmetry line x = 4. Setting x = 0 gives the y-intercept (0, 12).

These features constrain a sketch. An upward-opening curve with vertex above the x-axis would contradict the two computed roots. A sketch crossing at −2 and −6 would contradict substitution. The drawing should check the algebra, not merely decorate it.

A root is an input value; an intercept is a coordinate pair. State the requested object correctly. If the task asks to solve x² − 8x + 12 = 5, the desired inputs are intersections with y = 5, not necessarily the x-intercepts. Rearranging produces a different equation, x² − 8x + 7 = 0, with roots 1 and 7.

14. A model can reject a mathematically valid root

A rectangle has width w centimetres, length w + 3 centimetres and area 54 cm². The equation is w(w + 3) = 54, so w² + 3w − 54 = 0. Factorisation gives (w + 9)(w − 6) = 0.

The algebraic roots are −9 and 6. A width must be positive in this model, so only w = 6 is admitted. The length is 9 cm. Check both requirements: 6 × 9 = 54 and 9 − 6 = 3.

Do not say that −9 is an arithmetic mistake. It is a valid root of the polynomial equation but not a permitted physical dimension. This distinction helps learners separate algebraic correctness from contextual feasibility. Define the domain before solving whenever the variable counts objects, measures elapsed time or represents a physical length.

15. A border problem requires two changing dimensions

An invented rectangular panel measures 12 cm by 8 cm. A uniform border of width w is added outside all four sides. The total area becomes 192 cm². The outer dimensions are 12 + 2w and 8 + 2w, because the border adds width at both ends of each dimension.

Set (12 + 2w)(8 + 2w) = 192. Expansion gives 96 + 40w + 4w² = 192, hence w² + 10w − 24 = 0. The factors are (w + 12)(w − 2), so the non-negative width is 2 cm.

Check the outer rectangle: 16 × 12 = 192. Using 12 + w and 8 + w would describe a different border arrangement. The first failure in that incorrect solution would occur during modelling, before the quadratic method. A correct solution of the wrong equation is still not a solution of the stated geometry.

16. Restrictions can remove a candidate before or after factorisation

Solve (x² − 9)/(x − 3) = 0. First state x ≠ 3. A fraction is zero when its numerator is zero and its denominator is not. The numerator factorises to (x − 3)(x + 3), producing candidates 3 and −3. Only −3 is allowed.

Alternatively, simplify the fraction to x + 3 while retaining x ≠ 3, then solve x + 3 = 0. Both routes agree. The missing point in the original relationship is not restored by cancellation.

When a quadratic arises after squaring a more complicated equation, check every candidate in the unsquared original. Squaring preserves a true equality but can erase sign information. For √(x + 6) = x, the resulting quadratic has candidates 3 and −2, but only 3 satisfies the original. The root calculation is one stage of the argument, not always the final admission decision.

17. Recover an equation from its roots, but do not invent its scale

A monic quadratic with roots 2 and −5 is (x − 2)(x + 5) = 0, or x² + 3x − 10 = 0. Multiplying the entire equation by any non-zero constant preserves the same roots.

For a graph, however, that multiplier changes the vertical values. Knowing roots 2 and −5 gives y = a(x − 2)(x + 5), but does not determine a. If the graph also passes through (0, 20), then 20 = −10a, so a = −2.

This distinction is useful in reverse questions. Roots determine where the graph meets the axis, but not how high or low it lies elsewhere. Additional information is required. It also explains why two graphs can have the same roots without being identical functions.

18. Choose a method from the available structure

For x² = 49, take square roots. For x² − 9x + 20 = 0, factorisation is short. For (x + 3)² = 8, preserve the square. For x² + 6x + 2 = 0, completing the square gives exact roots directly. For 7x² − 4x − 9 = 0, the quadratic formula avoids an unproductive search for neat integer factors.

No single route is always best. The target can change the choice: a minimum-value question often benefits from completing the square, while a question about the number of roots may need only the discriminant.

A learner should explain the first choice in one sentence: “The square is already isolated”, “The integer factors are visible”, or “The question asks for a turning point”. That explanation is more useful than an automatic rule that every quadratic must be expanded and sent through the same formula.

19. Capstone: read one stipulated model in several ways

An invented height model is h(t) = −5t² + 20t + 1 for times t ≥ 0 while h is non-negative. Treat this as a stipulated mathematical model rather than measured motion. Find the initial height, greatest modelled height and first time the model reaches zero.

At t = 0, h = 1. Completing the square gives h(t) = −5(t − 2)² + 21. Since the squared term is non-negative and multiplied by −5, the maximum is 21 at t = 2.

For zero height, solve 5t² − 20t − 1 = 0. The formula gives t = [20 ± √420]/10 = 2 ± √105/5. The negative root is approximately −0.0494 and lies outside t ≥ 0. The admitted root is approximately 4.0494.

Check the broad shape: an initial positive height rises to its maximum at t = 2 and then falls to zero after t = 4. The useful domain ends at that zero because the example defines the model only while h is non-negative. Extending the polynomial further is possible algebraically, but is not automatically an extension of the stated situation.

20. Independent practice: twenty tasks

Work over the real numbers. Preserve exact forms unless a question requests rounding. For every equation, check all proposed roots and retain any original restrictions.

1. Factorise x² − 9x + 20, then solve the equation obtained by setting it equal to zero.
2. Solve x² + 2x − 24 = 0.
3. Solve 2x² + x − 6 = 0.
4. Solve 5x² − 20x = 0.
5. Solve 4x² − 49 = 0.
6. Solve (2x + 1)² = 9.
7. Solve (x − 4)² = 0.
8. Solve x² + 6x + 5 = 0 by completing the square.
9. Solve x² − 4x − 1 = 0 exactly.
10. Use the quadratic formula to solve 3x² + 2x − 5 = 0.
11. State the number of distinct real roots of x² + 2x + 5 = 0.
12. Find the vertex and roots of y = x² − 10x + 21.
13. Solve x(x − 1) = 12.
14. A rectangle has width w, length w + 4 and area 77. Find its positive dimensions.
15. Solve (x² − 16)/(x − 4) = 0, retaining the restriction.
16. A monic quadratic has roots −2 and 7. Write its equation.
17. Find a if y = a(x − 1)(x − 5) passes through (0, 10).
18. Solve x² + 5x = x² + 20.
19. Find the minimum value of 2x² − 12x + 23 and where it occurs.
20. An 8 by 6 rectangle gains an outside border of uniform width w. Total area becomes 120. Find w ≥ 0.

21. Worked answers with decisive checks

1. (x − 4)(x − 5); roots 4 and 5. The factors have product 20 and sum −9 in their constant terms. Setting each factor to zero finishes the equation-solving task. Substitution yields 16 − 36 + 20 = 0 and 25 − 45 + 20 = 0.

2. x = 4 or −6. Factorise to (x + 6)(x − 4) = 0. The opposite signs are required by the negative constant; the larger positive factor contributes the positive middle coefficient.

3. x = 3/2 or −2. The factors are (2x − 3)(x + 2). Solve both linear factors separately. At 3/2, the original expression is 9/2 + 3/2 − 6 = 0.

4. x = 0 or 4. Factorise as 5x(x − 4). Dividing by x before recording the zero case would lose the root 0. Both values make the original expression zero.

5. x = ±7/2. Use the difference of squares, (2x − 7)(2x + 7) = 0, or write x² = 49/4. Both signs are needed.

6. x = 1 or −2. Set 2x + 1 = 3 or −3. The original brackets become 3 and −3, and both square to 9.

7. x = 4. A square equals zero only when its base is zero. This is one distinct repeated root, not two different answers created by writing ±0.

8. x = −1 or −5. Rewriting gives (x + 3)² − 4 = 0. Thus x + 3 = ±2. Expand the completed square to check that the original constant 5 has been preserved.

9. x = 2 ± √5. Complete the square: (x − 2)² = 5. The two exact surd answers are appropriate; no rounding was requested.

10. x = 1 or −5/3. Here a = 3, b = 2, c = −5, so D = 4 + 60 = 64. The formula gives [−2 ± 8]/6. The whole numerator is divided by 6.

11. No real roots. D = 4 − 20 = −16. Equivalently, x² + 2x + 5 = (x + 1)² + 4, which cannot be zero for real x.

12. Vertex (5, −4); roots 3 and 7. The completed square is (x − 5)² − 4, and the factors are (x − 3)(x − 7). The midpoint of the two roots agrees with the symmetry line x = 5.

13. x = 4 or −3. Rearrange to x² − x − 12 = 0 before applying the zero-product property. The original products are 4(3) and (−3)(−4), both 12.

14. Width 7, length 11. Solve w² + 4w − 77 = (w + 11)(w − 7) = 0. Reject w = −11 because width is positive. Check 7 × 11 = 77 and difference 4.

15. x = −4 only, with x ≠ 4. The numerator vanishes at ±4, but 4 makes the original denominator zero. At −4, the denominator is −8 and the numerator is zero, so the fraction is defined and zero.

16. x² − 5x − 14 = 0. Multiply (x + 2)(x − 7). Monic specifies leading coefficient one; without it, non-zero multiples would share the same roots.

17. a = 2. Substitute the supplied point: 10 = a(−1)(−5) = 5a. The additional point fixes the scale that the roots alone could not determine.

18. x = 4. The x² terms cancel, leaving 5x = 20. The equation is linear after simplification; treating it as a quadratic with a = 0 in the formula would create an invalid denominator.

19. Minimum 5 at x = 3. Rewrite 2x² − 12x + 23 as 2(x − 3)² + 5. The square is non-negative and is zero at 3.

20. w = 2. Set (8 + 2w)(6 + 2w) = 120, giving 4w² + 28w − 72 = 0. Divide by 4 and factor (w + 9)(w − 2). Reject −9; the outer dimensions 12 and 10 give area 120.

22. Repair the decision that failed, not every method at once

A useful error record separates modelling, rearranging to zero, factorising, extracting roots, applying restrictions and checking. “Lost x = 0 by division” identifies a specific issue. “Bad at quadratics” does not. Likewise, a correct factorisation followed by only one root calls for a different repair from an incorrect factorisation.

Try a contrast pair after the correction. Compare x(x − 5) = 0 with x(x − 5) = 6. Compare √36 with solving x² = 36. Compare a bare polynomial with a rational expression carrying an excluded input. These pairs make the controlling condition change while much of the notation stays familiar.

For independent transfer, remove the method label and ask which route is shortest and why. The proposed routine is instructional guidance, not a guarantee of learning speed. Progress is evidenced by later unaided work, including the ability to explain why a method is legal and why all admitted roots survive the original question.

23. Continue through the BTT learning routes

Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Signed Numbers, Brackets and Algebraic Structure for the prerequisite expansions and factors.

Within Batch 05, Simultaneous Equations, Elimination and Modelling explains shared constraints; Vectors, Magnitude, Direction and Geometric Reasoning connects squared lengths with geometric quantities; and Quartiles, Cumulative Frequency, Box Plots and Standard Deviation gives a different use of squared deviations without turning that statistical calculation into a root-finding problem.

Sources and scope

For a standard reference on quadratic definitions and methods, consult OpenStax, College Algebra 2e, Section 2.5. The derivations, examples and twenty practice answers here are original. Their correctness is supported by the displayed calculations, not by an implied official endorsement.

The MOE secondary curriculum doorway and the learner’s school topic list should guide which methods and extensions to study. This article is a worked repair guide, not a claim that every section belongs to every secondary level or an official assessment specification.