Solving Quadratic Equations | Edexcel IGCSE Maths Revision Notes 2022 (2024)

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Solving Quadratics by Factorising

How do I solve a quadratic equation using factorisation?

Rearrange it into the form ax² + bx + c = 0
- zero must be on one side
  - it is easier to use the side where a is positive
Factorise the quadratic and solve each bracket equal to zero
- If (x + 4)(x - 1) = 0, then either x + 4 = 0 or x - 1 = 0
  - Because if A × B = 0, then either A = 0 or B = 0
Factorise the quadratic and solve each bracket equal to zero
To solve $(x - 3) (x + 7) = 0$
- …solve “first bracket = 0”:
  - x – 3 = 0
  - add 3 to both sides: x = 3
- …and solve “second bracket = 0”
  - x + 7 = 0
  - subtract 7 from both sides: x = -7
- The two solutions are x = 3 or x = -7
  - The solutions have the opposite signs to the numbers in the brackets
To solve $(2 x - 3) (3 x + 5) = 0$
- …solve “first bracket = 0”
  - 2x – 3 = 0
  - add 3 to both sides: 2x = 3
  - divide both sides by 2: x = $\frac{3}{2}$
- …solve “second bracket = 0”
  - 3x + 5 = 0
  - subtract 5 from both sides: 3x = -5
  - divide both sides by 3: x = $- \frac{5}{3}$
- The two solutions are x = $\frac{3}{2}$ or x = $- \frac{5}{3}$
To solve $x (x - 4) = 0$
- it may help to think of x as (x – 0) or (x)
- …solve “first bracket = 0”
  - (x) = 0, so x = 0
- …solve “second bracket = 0”
  - x – 4 = 0
  - add 4 to both sides: x = 4
- The two solutions are x = 0 or x = 4
  - It is a common mistake to divide both sides by x at the beginning -you will lose a solution (the x = 0 solution)

Exam Tip

Use a calculator to check your final solutions!
- Calculators also help you to factorise (if you're struggling with that step)
- A calculator gives solutions to $6 x^{2} + x - 2 = 0$ as x = $- \frac{2}{3}$ and x = $\frac{1}{2}$
  - "Reverse" the method above to factorise!
  - $6 x^{2} + x - 2 = (3 x + 2) (2 x - 1)$
- Warning: a calculator gives solutions to 12x² + 2x – 4 = 0 as x = $- \frac{2}{3}$ and x = $\frac{1}{2}$
  - But 12x² + 2x – 4 ≠ $(3 x + 2) (2 x - 1)$ as these brackets expand to 6x² + ... not 12x² + ...
  - Multiply by 2 to correct this
  - 12x² + 2x – 4 = $2 (3 x + 2) (2 x - 1)$

Worked example

(a)

Solve $(x - 2) (x + 5) = 0$

Set the first bracket equal to zero

x – 2 = 0

Add 2 to both sides

x = 2

Set the second bracket equal to zero

x + 5 = 0

Solving by Completing the Square

How do I solve a quadratic equation by completing the square?

To solvex² + bx + c = 0
- replace the first two terms, x² + bx, with (x + p)² - p² where p is half of b
- this is called completing the square
  - x² + bx + c = 0 becomes
    - (x + p)² - p² + c = 0 where p is half of b
- rearrange this equation to make x the subject (using ±√)
For example, solve x² + 10x + 9 = 0 by completing the square
- x² + 10x becomes(x + 5)² - 5²
- so x² + 10x + 9 = 0 becomes(x + 5)² - 5²+ 9 = 0
- make x the subject (using ±√)
  - (x + 5)² - 25+ 9 = 0
  - (x + 5)² = 16
  - x + 5 = ±√16
  - x = ±4 - 5
  - x = -1 orx = -9

If the equation is ax² + bx + c = 0 with a number in front of x², then divide both sides by a first, before completing the square

How does completing the square link to the quadratic formula?

The quadratic formula actually comes from completing the square to solve ax² + bx + c = 0
- a, b and c are left as letters, to be as general as possible
You can see hints of this when you solve quadratics
- For example, solving x² + 10x + 9 = 0
  - by completing the square,(x + 5)² = 16 sox = ± 4 - 5 (from above)
  - by the quadratic formula, $x = \frac{- 10 \pm \sqrt{64}}{2} = - 5 \pm \frac{8}{2}$ = ± 4 - 5 (the same structure)

Exam Tip

When making x the subject to find the solutions at the end, don't expand the squared brackets back out again!
- Remember to use ±√ to get two solutions

Worked example

Solve $2 x^{2} - 8 x - 24 = 0$ by completing the square

Divide both sides by 2 to make the quadratic start with x²

$x^{2} - 4 x - 12 = 0$

Halve the middle number, -4, to get -2
Replace the first two terms, x² - 4x, with (x - 2)² - (-2)²

${(x - 2)}^{2} - {(- 2)}^{2} - 12 = 0$

Simplify the numbers

$\begin{array}{rcl} {(x - 2)}^{2} - 4 - 12 & = & 0 \\ {(x - 2)}^{2} - 16 & = & 0 \end{array}$

Add 16 to both sides

${(x - 2)}^{2} = 16$

Square root both sides
Include the ± sign to get two solutions

$x - 2 = \pm \sqrt{16} = \pm 4$

Add 2 to both sides

$x = \pm 4 + 2$

Work out each solution separately

x = 6 or x = -2

Quadratic Formula

How do I use the quadratic formula to solve a quadratic equation?

A quadratic equation has the form:
ax² + bx + c = 0 (as long as a ≠ 0)
- you need "= 0" on one side
The quadratic formula is a formula that gives both solutions:
- $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Read off the values of a, b and c from the equation
Substitute theseinto the formula
- write this line of working in the exam
- Put brackets around any negative numbers being substituted in
To solve 2x² - 7x - 3 = 0 using the quadratic formula:
- a = 2, b = -7 and c = -3
- $x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 \times 2 \times (- 3)}}{2 \times 2}$
- Type this into a calculator
  - once with + for ± and once with - for ±
- The solutions are x = 3.886 and x = -0.386 (to 3 dp)
  - Rounding is often asked for in the question
  - The calculator also gives these solutions in exact form (surd form), if required
  - x = $\frac{7 + \sqrt{73}}{4}$ and x = $\frac{7 - \sqrt{73}}{4}$

What is the discriminant?

The part of the formula under the square root (b² – 4ac) is called the discriminant
The sign of this value tells you if there are 0, 1 or 2 solutions
- If b² – 4ac > 0 (positive)
  - then there are 2 different solutions
- If b² – 4ac = 0
  - then there is only 1 solution
  - sometimes called "two repeated solutions"
- If b² – 4ac < 0 (negative)
  - then there are no solutions
  - If your calculator gives you solutions with i terms in, these are "complex" and not what we are looking for
- Interestingly, if b² – 4ac is a perfect square number ( 1, 4, 9, 16, …) then the quadratic expression could have been factorised!

Can I use my calculator to solve quadratic equations?

Yes tocheck your final answers, but a method must still be shown as above

Exam Tip

Make sure the quadratic equation has "= 0" on the right-hand side, otherwise it needs rearranging first
Always look for how the question wants you to leave your final answers
- for example, correct to 2 decimal places

Worked example

Use the quadratic formula to find the solutions of the equation 3x² - 2x - 4 = 0, giving your answers correct to 3 significant figures.

Write down the values of a, b and c

a = 3, b = -2, c = -4

Substitute these values into the quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Put brackets around any negative numbers

$x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 \times 3 \times (- 4)}}{2 \times 3}$

Input this into a calculator
Use + for ± to get the first solution

x = 1.53518...

Input this into a calculator a second time
Use - for ± to get the second solution

x = 0.86851...

Present both answers together (using the word "or" between them)
Round the answers correct to 3 significant figures (note how this affects the number of decimal places)

x = 1.54 or x = 0.869

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