Choosing a method is often the first difficulty in a quadratic equation. Some expressions fall into two brackets immediately. Others become easier when you make a perfect square, while the quadratic formula gives a reliable route when simple factors are hard to see. This revision guide works through each choice, explains the signs, and ends with questions you can solve before checking the answers.
The questions below are original practice examples. They are intended to build mathematical understanding for KCSE revision; they are not reproduced examination questions or an official marking scheme.
Start by putting the equation in a useful form
A quadratic equation can be written as ax2 + bx + c = 0, where a is not zero. The highest power of the unknown is two. Before choosing a method, expand any brackets, collect like terms and move everything to one side. Record a, b and c with their signs. A missing term has coefficient zero.
For example, 3x2 + 5 = 2x becomes 3x2 − 2x + 5 = 0. Here a = 3, b = −2 and c = 5. Writing b = 2 changes the equation and produces the wrong answer. If every term has a common numerical factor, dividing the whole equation by that factor can make the work shorter.
A practical method-choice checklist
- Look for a common factor first. In x2 − 6x = 0, taking out x is quicker than searching for two unrelated brackets.
- Try factorising when you can identify suitable integer or simple fractional roots without a long search.
- Use completing the square when the coefficient of x2 is one, or becomes one easily, and you want an exact expression or the turning-point form.
- Use the formula when factors are inconvenient, or when the question specifically requests it.
- Follow a stated method in the question even when you know another route.
These methods express the same underlying algebra. The OpenStax explanation of quadratic equations also shows how completing the square leads to the general formula.
Example 1: factorising a quadratic with a leading coefficient of one
Solve x2 − 7x + 12 = 0. We need two numbers whose product is 12 and whose sum is −7. Both numbers must be negative because their product is positive and their sum is negative. The pair is −3 and −4.
x2 − 7x + 12 = (x − 3)(x − 4).
Therefore (x − 3)(x − 4) = 0. A product is zero when at least one of its factors is zero, so x − 3 = 0 or x − 4 = 0. The solutions are x = 3 or x = 4.
Check both roots in the original equation. At x = 3, the left side is 9 − 21 + 12 = 0. At x = 4, it is 16 − 28 + 12 = 0. This quick substitution also catches accidental sign reversals in the brackets.
Example 2: factorising when the first coefficient is larger
Solve 6x2 + 7x − 3 = 0. Multiply the first and last coefficients: 6 × (−3) = −18. Find two numbers that multiply to −18 and add to 7. They are 9 and −2.
- Split the middle term: 6x2 + 9x − 2x − 3 = 0.
- Group: 3x(2x + 3) − 1(2x + 3) = 0.
- Take out the shared bracket: (3x − 1)(2x + 3) = 0.
- Solve each linear equation: x = 1/3 or x = −3/2.
Expand your brackets as a separate check: 6x2 + 9x − 2x − 3 returns the original expression. Splitting the middle term is useful because it replaces a guess with a clear sequence of operations.
Example 3: completing the square when factors are awkward
Solve x2 − 6x + 2 = 0. Move the constant: x2 − 6x = −2. Half of the x coefficient is −3, and its square is 9. Add 9 to both sides:
x2 − 6x + 9 = 7, so (x − 3)2 = 7.
Taking square roots gives x − 3 = ±√7. Hence x = 3 + √7 or x = 3 − √7. To three decimal places, these are approximately 5.646 and 0.354.
Keep the exact roots unless a decimal answer is requested. The plus-or-minus sign matters: two different numbers have square 7. Also notice that adding 9 to only the left side would change the equation. Completing a square must preserve equality.
Example 4: using the quadratic formula carefully
Solve 2x2 − 4x − 3 = 0. The formula is x = [−b ± √(b2 − 4ac)] / (2a). Here a = 2, b = −4 and c = −3.
The expression under the square root is (−4)2 − 4(2)(−3) = 16 + 24 = 40. Therefore x = [4 ± √40] / 4. Since √40 = 2√10, the exact answers are x = 1 ± √10/2. The decimal answers are approximately 2.581 and −0.581.
Use brackets around the complete numerator when entering the calculation. Dividing only the square-root part by 4 gives a different value. Retain the full calculator result until the final rounding step, particularly when a later part of the problem uses the root.
Read the discriminant before doing unnecessary work
The expression D = b2 − 4ac is the discriminant. For real coefficients, D greater than zero gives two distinct real roots; D equal to zero gives one repeated real root; and D less than zero gives no real roots. See the OpenStax quadratic terminology reference for the associated definitions.
For 4x2 − 12x + 9 = 0, D = 144 − 144 = 0, so x = 12/8 = 3/2 is repeated. For x2 + 2x + 5 = 0, D = 4 − 20 = −16. In a question restricted to real numbers, the appropriate conclusion is that there are no real solutions.
Example 5: build an equation from a rectangle
A rectangular display board has area 84 cm2. Its length is 5 cm greater than its width. Find its dimensions.
Let the width be w cm. The length is w + 5 cm, so w(w + 5) = 84. Rearranging gives w2 + 5w − 84 = 0. Factorising produces (w + 12)(w − 7) = 0. Thus w = −12 or w = 7.
The negative value solves the algebraic equation but cannot describe this board’s width. The dimensions are 7 cm by 12 cm. Check both original conditions: 7 × 12 = 84 and 12 − 7 = 5. A complete applied answer explains why an inadmissible root is rejected.
Find the first wrong step, not just the wrong answer
Consider this proposed solution to 2x2 − 8x = 0: divide by 2x, giving x − 4 = 0, so x = 4. The value 4 is a solution, but the method has silently assumed x is not zero. It loses another valid solution.
Instead, factorise: 2x(x − 4) = 0. Either x = 0 or x = 4. Substitution confirms both. Record the correction as “factor out the variable before applying the zero-product rule”, rather than “remember a second answer”. Some quadratics have a repeated root, so merely forcing every question to have two different answers would create a new mistake.
For a fresh check, solve 3y2 + 12y = 0. The factorisation is 3y(y + 4) = 0, giving y = 0 or −4. Explain why dividing by y at the start would be unsafe.
Try these questions before opening the solutions
- Solve x2 + x − 20 = 0 by factorising.
- Solve 3x2 − 5x − 2 = 0 and leave both answers exact.
- Complete the square to solve x2 + 8x + 3 = 0.
- Use the formula for 2x2 + 3x − 4 = 0.
Worked answer check
1. (x + 5)(x − 4) = 0, so x = −5 or 4. The two numbers have product −20 and sum 1.
2. (3x + 1)(x − 2) = 0, giving x = −1/3 or 2. Expanding gives 3x2 − 6x + x − 2.
3. (x + 4)2 = 13, so x = −4 ± √13. The number added on each side is 16.
4. D = 9 + 32 = 41, so x = (−3 ± √41)/4, approximately 0.851 or −2.351.
Questions learners ask about quadratic methods
Can I always use the formula?
It applies to every quadratic with a nonzero leading coefficient. A question may still require factorisation or completing the square, so read the instruction first.
Why must one side be zero when factorising?
The zero-product rule works when the product equals zero. From (x − 2)(x + 1) = 6, you cannot set either bracket equal to zero.
What happens if c is zero?
Take out x. For x2 − 6x = 0, write x(x − 6) = 0 and obtain x = 0 or 6. Dividing by x would lose the zero solution.
Does every quadratic have two different answers?
No. A quadratic can have a repeated root or no real roots. Use its discriminant to distinguish these cases.
Can a fraction be a correct root?
Yes. Roots do not have to be whole numbers. Substitute the fraction into the original equation to check it without introducing rounding errors.
Should I round a square root immediately?
Keep it exact while working. Round at the end to the precision requested, or retain a surd when no decimal approximation is required.
Why do I halve the middle coefficient?
Expanding (x + k)2 gives x2 + 2kx + k2. Matching 2k to the middle coefficient determines the number inside the bracket.
How should I revise after making an error?
Identify the first incorrect line, explain the rule it broke, and solve one similar question from a blank page. Checking only the final root can hide the underlying mistake.
Continue with focused algebra practice
Once you can explain your method choices, visit the KCSE Mathematics topic-tests page for the site’s existing subject revision destination. That page currently asks visitors to log in. For other subjects, use the KCSE topic-test directory. If you need account access, review the current information on the membership page before deciding whether to join.
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