Solving Quadratic Equations — Grade 11 Maths (CAPS)

All the ways to solve a quadratic — factorising, completing the square, the formula, rearranging first, a leading coefficient, and quadratics hidden in fractions — plus a worked example for each and a full worksheet.

A quadratic equation has the form ax² + bx + c = 0 (highest power 2). This guide covers all three methods and the ways Grade 11 exams combine them. The order to try: (1) get one side = 0, (2) factorise if you can, (3) otherwise use the formula — unless the question tells you to complete the square, which is a method in its own right and is where the formula comes from.

1Method 1: Factorising (the zero-product rule)

If a product equals zero, at least one factor must be zero. So factorise, set each bracket to zero, and solve.

Worked ExampleExample 1: x² − 5x + 6 = 0

1
Factorise: two numbers that multiply to 6, add to −5 → −2 and −3 → (x − 2)(x − 3) = 0.
Standard trinomial factorising.
2
Set each bracket to zero: x − 2 = 0 or x − 3 = 0.
Zero-product rule.
3
x = 2 or x = 3.
A quadratic usually has two solutions.

Worked ExampleExample 2: common factor: x² − 7x = 0

1
Take out x: x(x − 7) = 0.
Don't divide by x — you'd lose the root x = 0.
2
x = 0 or x − 7 = 0.
Zero-product rule.
3
x = 0 or x = 7.
Both are valid roots.

2Method 2: The quadratic formula

When it won't factorise, use the formula (on your data sheet). It works for every quadratic.

x = (−b ± √(b² − 4ac))/2a (a ≠ 0)

Worked ExampleExample 3: 2x² + 3x − 1 = 0 (won't factorise)

1
a = 2, b = 3, c = −1.
Match to ax² + bx + c = 0.
2
x = (−3 ± √(9 + 8))/4 = (−3 ± √17)/4.
b² − 4ac = 9 − 4(2)(−1) = 17.
3
x ≈ 0,28 or x ≈ −1,78.
Work out both the + and − versions; round to 2 decimals.

3Method 3: Completing the square

The ATP puts this one first in Grade 11 — “1. Complete the square” — and for good reason: it is where the quadratic formula itself comes from, it is what the turning point of a parabola is found with, and an exam will sometimes instruct you to use it. The idea is to force the left side into a perfect square, so that one square root finishes the job.

1. Get x² alone in front (divide by a if it is not 1).
2. Move the constant to the right.
3. Halve the coefficient of x, square it, and add it to both sides.
4. Write the left side as (x + p)² and take the square root — remembering ±.

Worked ExampleExample 4: solve x² + 6x + 5 = 0 by completing the square

1
x² + 6x = −5
Move the constant across first.
2
Half of 6 is 3, and 3² = 9. Add 9 to BOTH sides: x² + 6x + 9 = −5 + 9 = 4
Adding to one side only changes the equation.
3
(x + 3)² = 4
The left side is now a perfect square, and the 3 is the half you took.
4
x + 3 = ±2
Square-root both sides — both signs, or you lose a root.
5
x = −1 or x = −5
Check by factorising: (x + 1)(x + 5) = 0 ✓

Worked ExampleExample 5: solve 2x² + 8x + 3 = 0, correct to TWO decimal places

1
Divide every term by 2: x² + 4x + 1,5 = 0
The method needs a coefficient of 1 in front of x². Divide the WHOLE equation, not just two terms.
2
x² + 4x = −1,5
Constant to the right.
3
Half of 4 is 2, and 2² = 4: x² + 4x + 4 = −1,5 + 4 = 2,5
Add 4 to both sides.
4
(x + 2)² = 2,5, so x + 2 = ±√2,5
√2,5 ≈ 1,5811 — keep it in the calculator.
5
x ≈ −0,42 or x ≈ −3,58
The formula gives the same two roots: b² − 4ac = 64 − 24 = 40 ✓

💡

The same four steps rewrite an expression in the form (x + p)² + q without solving anything — and that form hands you the turning point of the parabola y = (x + p)² + q straight away. Same work, two different questions.

⚠️

Two marks are lost here more than anywhere else. Adding the square to one side only — whatever you add on the left must be added on the right. And dropping the ± when you take the square root, which throws away one of the two roots. A third: with 2x² + 8x + 3 = 0 you must divide the whole equation by 2, the 3 included.

4Type 4: Rearrange to standard form first

Worked ExampleExample 6: x² = 3x + 10

1
Move everything to one side: x² − 3x − 10 = 0.
The equation must equal 0 before solving.
2
Factorise: (x − 5)(x + 2) = 0.
Multiply to −10, add to −3 → −5 and +2.
3
x = 5 or x = −2.

5Type 5: A leading coefficient

Worked ExampleExample 7: 3x² − 10x + 8 = 0

1
ac = 24; two numbers multiply to 24, add to −10 → −6 and −4.
The ac-method for factorising when a ≠ 1.
2
Split: 3x² − 6x − 4x + 8 = 3x(x − 2) − 4(x − 2) = (x − 2)(3x − 4).
Factorise by grouping.
3
x = 2 or 3x − 4 = 0 → x = 4/3.

6Type 6: Quadratics hidden in fractions

Worked ExampleExample 8: x + 6/x = 5

1
Multiply every term by x: x² + 6 = 5x.
Clear the fraction (note x ≠ 0).
2
Rearrange: x² − 5x + 6 = 0.
Standard form.
3
(x − 2)(x − 3) = 0 → x = 2 or x = 3.
Both are non-zero, so both are valid.

💡

Always make one side zero first, then try factorising before the formula. Do not divide both sides by x when x could be 0 — you lose the root x = 0; take x out as a common factor instead.

⚠️

Don't lose a solution. A quadratic normally has two answers — writing only one is a common way to drop marks. And check any restrictions (like x ≠ 0) when you multiplied out a fraction.

Quadratics run through functions, calculus and geometry, so fluency here pays off everywhere. Work every type in the worksheet.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Solve x² − 7x + 12 = 0. (factorise)
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(x − 3)(x − 4)=0 → x = 3 or 4
2
Solve x² + 2x − 8 = 0. (factorise)
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(x + 4)(x − 2)=0 → x = −4 or 2
3
Solve x² − 9x = 0. (common factor)
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x(x − 9)=0 → x = 0 or 9
4
Solve x² − 16 = 0. (difference of squares)
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(x − 4)(x + 4)=0 → x = ±4
5
Solve x² + 4x + 1 = 0 (formula, 2 decimal places). (formula)
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x = (−4±√12)/2 ≈ −0,27 or −3,73
6
Solve 2x² − 5x + 1 = 0 (formula, 2 decimal places). (formula)
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x = (5±√17)/4 ≈ 2,28 or 0,22
7
Solve x² + 8x − 9 = 0 by completing the square. (complete the square)
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x² + 8x = 9; half of 8 is 4 and 4² = 16, so x² + 8x + 16 = 9 + 16 = 25; (x + 4)² = 25; x + 4 = ±5 → x = 1 or x = −9
8
Solve 3x² − 12x + 5 = 0 by completing the square, correct to TWO decimal places. (complete the square)
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÷ 3: x² − 4x + 5/3 = 0 → x² − 4x = −5/3; add 4: (x − 2)² = 7/3; x − 2 = ±1,5275 → x ≈ 3,53 or x ≈ 0,47
9
Write x² − 10x + 3 in the form (x + p)² + q. (complete the square)
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Half of −10 is −5 and (−5)² = 25, so x² − 10x + 3 = (x − 5)² − 25 + 3 = (x − 5)² − 22
10
Solve x² = 2x + 15. (rearrange)
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x² − 2x − 15=0 → (x − 5)(x + 3)=0 → x = 5 or −3
11
Solve 2x² + 7x + 3 = 0. (leading coeff)
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(2x + 1)(x + 3)=0 → x = −1/2 or −3
12
Solve x + 5/x = 6. (fraction)
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x² − 6x + 5=0 → (x − 1)(x − 5)=0 → x = 1 or 5
13
Solve (x − 2)(x + 1) = 4. (expand first)
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x² − x − 2=4 → x² − x − 6=0 → (x − 3)(x + 2)=0 → x = 3 or −2