WAEC / WASSCE
WAEC Quadratic Equations Practice Questions, Fully Solved
Real WAEC past papers are copyrighted, so this page gives you the next best thing: six original questions written in the exact WASSCE format, covering factorisation, the formula, completing the square, word problems, roots relationships and graph reading. Each is fully solved with WAEC-style M (method) and A (accuracy) mark tips.
Most pages ranking for WAEC quadratic questions are scanned PDFs on document-sharing sites, half of them behind download walls. This page does it differently. Because real past papers are copyrighted, I wrote six original questions matched to the WASSCE format in topic, difficulty and structure, then solved each in full with the marking logic shown. One honesty note: I sit Cambridge IGCSE and A-Levels, not WAEC, so this comes from studying the WASSCE General Mathematics syllabus, paper structure and chief examiners' reports, not from sitting the paper myself. Work each question before reading the solution, then check where the M and A marks would have landed in your attempt.
How quadratics appear in Paper 1 vs Paper 2
WASSCE General Mathematics has two written papers, and quadratics behave differently in each. Paper 1 is 50 objective questions in 1 hour 30 minutes. Quadratic items there are short: pick the roots of a factorisable equation, spot the equation whose roots are given, or find a sum or product of roots. Speed matters, so factorisation and the sum-and-product results are the tools to drill. Paper 2 lasts 2 hours 30 minutes and splits into Section A (five compulsory questions) and Section B (choose five from ten). Quadratics in Paper 2 come as multi-part questions: solve by completing the square or the formula to a stated number of decimal places, form an equation from a word problem, or build a table of values, plot the curve and read solutions off the graph. The graph question is a popular Section B choice because a large slice of its marks sits in the table and the plot rather than in algebra.
How WAEC marks a quadratic question
WAEC marking schemes label most marks M or A. An M mark is for method: substituting into the quadratic formula, a correct factorisation attempt, adding the right constant to both sides when completing the square. You keep it even if a later arithmetic slip spoils the answer. An A mark is for accuracy: the correct value, in the demanded form, following correct method. A marks usually cannot be earned without the M mark in front of them, which is why working matters even when you can see the answer in your head. Some schemes also award B marks for an independent correct statement, such as roots written straight from a graph. The practical consequences: quote the formula before substituting into it, show the factor pair you tried, state rejected roots and why you rejected them, and round only at the final line. A bare answer scores fully in Paper 1, where only the shaded option counts, but in Paper 2 it collects one mark at best.
Calculator rules in WASSCE maths
WAEC allows silent, cordless, non-programmable calculators in both mathematics papers, and phones are banned outright, even as clocks. So a calculator handles sqrt(41) in a formula question with no trouble. Two traps come with that permission. First, when a question says give your answers correct to two decimal places, carry at least three or four decimal places through the working and round once at the end; rounding a square root early gives answers a hundredth off and costs the A mark. Second, when a question says leave your answer in surd form, a decimal is wrong even though the calculator offers one: 3 + sqrt(5) is the answer, 5.24 is not. Many scientific calculators now solve quadratics outright in an equation mode. Use that to check your roots, never to replace working, because in Paper 2 the marks are attached to visible method, not to the final number.
Worked questions, step by step
Solve the equation x^2 - 5x - 14 = 0.
- Look for two numbers that multiply to -14 and add to -5: they are -7 and +2. Factorising needs exactly this pair.
- Write x^2 - 5x - 14 = (x - 7)(x + 2) = 0. Expanding the brackets back confirms the factorisation.
- Set each bracket equal to zero: x - 7 = 0 or x + 2 = 0. A product is zero only when one of its factors is zero.
- Solve each: x = 7 or x = -2. Both values satisfy the original equation.
Answer: x = 7 or x = -2
Where marks slip: The M mark here is for a visible factorisation, so writing only the two answers with no brackets shown risks losing everything if either root is wrong.
Try one yourself: Solve x^2 - 3x - 10 = 0 by factorisation. (Answer: x = 5 or x = -2)
Using the quadratic formula, solve 2x^2 + 3x - 4 = 0, giving your answers correct to two decimal places.
- Compare with ax^2 + bx + c = 0 to get a = 2, b = 3, c = -4. Identifying a, b and c correctly is the first method step.
- Quote the formula: x = (-b + or - sqrt(b^2 - 4ac)) / (2a). Writing it before substituting lets the marker follow your work.
- Substitute: x = (-3 + or - sqrt(9 - 4(2)(-4))) / 4 = (-3 + or - sqrt(41)) / 4. The discriminant is 9 + 32 = 41 because subtracting a negative adds.
- Evaluate sqrt(41) = 6.403 to three decimal places. Keep an extra decimal place so the final rounding is safe.
- Compute both roots: x = (-3 + 6.403)/4 = 0.85 and x = (-3 - 6.403)/4 = -2.35, each to two decimal places.
Answer: x = 0.85 or x = -2.35 (to 2 decimal places)
Where marks slip: There is an M mark for correct substitution and an A mark for each root, and rounding sqrt(41) too early is the single most common way candidates lose both A marks.
Try one yourself: Use the quadratic formula to solve 3x^2 - 5x - 1 = 0, correct to two decimal places. (Answer: x = 1.85 or x = -0.18)
The product of two consecutive positive odd numbers is 195. Find the two numbers.
- Let the smaller number be n, so the next odd number is n + 2. Defining the variable in words earns the setup mark.
- Form the equation n(n + 2) = 195, which gives n^2 + 2n - 195 = 0. Expand and bring every term to one side before factorising.
- Factorise: (n + 15)(n - 13) = 0, since 15 and -13 multiply to -195 and add to 2.
- So n = -15 or n = 13, and n = -15 is rejected because the numbers are positive. Write the rejection down rather than silently dropping it.
- The numbers are 13 and 15, and checking gives 13 x 15 = 195. A one-line check protects the final answer mark.
Answer: 13 and 15
Where marks slip: Chief examiners repeatedly note candidates leaving both roots standing; the mark for rejecting n = -15 with a reason is only given when the rejection is written.
Try one yourself: The product of two consecutive positive even numbers is 168. Find the two numbers. (Answer: 12 and 14)
Solve x^2 - 6x + 4 = 0 by completing the square, leaving your answer in surd form.
- Move the constant across: x^2 - 6x = -4. Completing the square works on the x terms alone.
- Halve the coefficient of x to get -3, then add (-3)^2 = 9 to both sides: x^2 - 6x + 9 = 5. Adding to both sides keeps the equation balanced.
- Write the left side as a perfect square: (x - 3)^2 = 5. This is the completed-square form the M marks are attached to.
- Take square roots of both sides: x - 3 = sqrt(5) or x - 3 = -sqrt(5). The negative root must appear or one solution is lost.
- Add 3 to both: x = 3 + sqrt(5) or x = 3 - sqrt(5). Surd form means leave sqrt(5) exactly as it is.
Answer: x = 3 + sqrt(5) or x = 3 - sqrt(5)
Where marks slip: Two habits lose marks on exactly this question: forgetting the negative square root (one A mark gone) and converting to decimals when surd form was demanded (both A marks gone).
Try one yourself: Solve x^2 + 4x - 3 = 0 by completing the square, leaving your answer in surd form. (Answer: x = -2 + sqrt(7) or x = -2 - sqrt(7))
If alpha and beta are the roots of the equation 3x^2 - 5x - 2 = 0, find the values of (a) alpha + beta, (b) alpha x beta, (c) 1/alpha + 1/beta.
- For ax^2 + bx + c = 0, the sum of roots is -b/a and the product is c/a. Quote these standard results before using them.
- Here a = 3, b = -5, c = -2, so alpha + beta = -(-5)/3 = 5/3. The two minus signs cancel.
- The product is alpha x beta = c/a = -2/3. Keep the negative sign with the fraction.
- Combine the reciprocals over a common denominator: 1/alpha + 1/beta = (alpha + beta)/(alpha x beta). This identity turns part (c) into pure substitution.
- Substitute: (5/3) divided by (-2/3) = 5/3 x 3/(-2) = -5/2. Dividing by a fraction means multiplying by its reciprocal.
Answer: (a) alpha + beta = 5/3 (b) alpha x beta = -2/3 (c) 1/alpha + 1/beta = -5/2
Where marks slip: The classic slip is a sign error in -b/a when b is already negative, and it forfeits the accuracy marks in every later part, so check part (a) twice.
Try one yourself: If alpha and beta are the roots of 2x^2 - 3x - 5 = 0, find 1/alpha + 1/beta. (Answer: -3/5)
The graph of y = x^2 - 2x - 8 is drawn for values of x from -4 to 6. The curve cuts the x-axis at x = -2 and at x = 4, and has a minimum turning point. (a) Write down the roots of x^2 - 2x - 8 = 0. (b) Find the coordinates of the minimum point. (c) Explain how the same graph can be used to solve x^2 - 2x - 8 = 7, and state the solutions.
- (a) The roots are the x-values where the curve cuts the x-axis, so x = -2 and x = 4. Graphically, a root is a crossing point on the x-axis.
- (b) By symmetry the minimum lies midway between the roots: x = (-2 + 4)/2 = 1. A parabola is symmetric about its turning point.
- Substitute x = 1 into the equation: y = 1 - 2 - 8 = -9, so the minimum point is (1, -9). Substituting the axis of symmetry gives the least value of y.
- (c) Draw the horizontal line y = 7 on the same axes and read off the x-values where it crosses the curve. The equation x^2 - 2x - 8 = 7 asks where the curve has height 7.
- Checking by algebra: x^2 - 2x - 15 = 0 factorises to (x - 5)(x + 3) = 0, so x = 5 or x = -3. The graph readings should match these values.
Answer: (a) x = -2 and x = 4 (b) minimum point (1, -9) (c) draw the line y = 7 and read the crossings: x = 5 or x = -3
Where marks slip: In WAEC graph questions there is usually a mark for the drawn line y = 7 itself, so candidates who solve part (c) purely by algebra without touching the graph leave that mark on the table.
Try one yourself: The graph of y = x^2 - 4x - 5 cuts the x-axis at x = -1 and x = 5. Find the minimum point and use the graph method to solve x^2 - 4x - 5 = 7. (Answer: minimum (2, -9); x = 6 or x = -2)
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Where can I find real WAEC past questions on quadratic equations?
The official route is WAEC's own e-learning portal, which hosts a selection of past questions with chief examiner commentary, and licensed printed past-question books are sold across West Africa. Random PDF and document-site uploads are usually incomplete scans. This page takes a different approach: original questions in the exact WASSCE format, which is legal to publish and just as useful for practice.
Which quadratic techniques does WASSCE actually examine?
The syllabus expects solving by factorisation, completing the square and the formula, forming quadratic equations from word problems, sum and product of roots, and graphical solution, including reading roots from a plotted curve. Simultaneous equations where one equation is quadratic also appear. The six questions above cover each of these at the difficulty the real papers use.
Can I lose marks for using the formula when the question says factorise?
Yes. When a question names a method, the method marks are tied to that method: solve by completing the square means the M marks are for completing-the-square steps, and a formula solution earns the answer marks at most. When the question just says solve, any valid method scores fully, so read the command words before starting.
How many marks is a quadratic question worth in Paper 2?
WAEC does not fix a single value. A quadratic part inside a compulsory Section A question commonly carries around 4 to 6 marks, while a Section B graph question spreads its marks across the table of values, the plotted curve, any drawn line and the readings. Whatever the total, the method and accuracy split follows the pattern shown in the six solutions above.