Learning Goal

Syllabus Requirement

Students need to learn the technique of completing the square and the solution of quadratic equations by:

  •    Factorisation.
  •    Use of the formula.
  •    Completing the square.

Study Notes

There are 3 different ways to solve quadratic functions and you should be familiar with them all, as in the exam you may be asked to use a specific method. The methods are the quadratic equation, completing the square and factorisation and they are all explained in detail below.

The 3 Methods for Solving a Quadratic Function

Method 1: Using the formula
Recall that for a quadratic function of the form ??ax^{2} + bx + c = 0?? the roots (solutions) can be found by using the formula: $$x = \frac{-b\pm \sqrt{b^{2} - 4ac}}{2a}$$ Method 2: Completing the Square
Completing the square allows you to solve a quadratic function, ??ax^{2} + bx + c = 0??:
  1. First, we need to ensure that ??a = 1?? in our quadratic function, if it does not divide all of the terms in the quadratic function by ??a?? to ensure that ??a = 1??, from now on the terms ??b?? and ??c?? should be taken from the quadratic function where ??a = 1??.
  2. Completing the square relies on knowing that: $$x{^2} + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^{2}$$
  3. To complete the square on the left hand side we need to add ??\left(\frac{b}{2}\right)^{2}??, which we must also add to the right hand side as well to keep the equation valid, we can also move the ??c?? to the right hand side: $$x^{2} + bx + c = 0$$ $$\implies x^{2} + bx = - c$$ $$\implies x^{2} + bx + \left(\frac{b}{2}\right)^{2} = -c + \left(\frac{b}{2}\right)^{2}$$
  4. Next complete the square on the left hand side: $$x^{2} + bx + \left(\frac{b}{2}\right)^{2} = -c + \left(\frac{b}{2}\right)^{2}$$ $$\implies \left(x + \frac{b}{2}\right)^{2} = -c + \left(\frac{b}{2}\right)^{2}$$
  5. Now take the square root of both sides: $$\left(x + \frac{b}{2}\right)^{2} = -c + \left(\frac{b}{2}\right)^{2}$$ $$\implies x + \frac{b}{2} = \pm \sqrt{-c + \left(\frac{b}{2}\right)^{2}}$$
  6. Finally, rearrange for ??x??: $$x + \frac{b}{2} = \pm \sqrt{-c + \left(\frac{b}{2}\right)^{2}}$$ $$\implies x = -\frac{b}{2} \pm \sqrt{-c + \left(\frac{b}{2}\right)^{2}} $$


Method 3: Factorisation
Not all quadratic equations can be factorised, but when they can be it is usually the most straightforward and error-free method for solving an equation of the form ??ax^{2} + bx + c = 0??. First factorise into the product of two linears: $$ax^{2} + bx + c =(dx+e)(fx+g)$$
Now since ??(dx+e)(fx+g)=0?? it must be the case that either ??(dx+e)=0?? or ??(fx+g)=0??.

When ??(dx+e)=0?? $$dx=-e \implies x=-\frac{e}{d}$$
When ??(fx+g)=0?? $$fx=-g \implies x=-\frac{g}{f}$$

In equations where the coefficient of ??x^2?? is equal to 1, factorisation is simpler: $$x^{2} + bx + c = (x + e)(x + g) = 0$$ Now $$(x+e)=0 \implies x=-e$$ and $$(x+g)=0 \implies x=-g$$


Worked Examples

We'll now run through some examples that cover each of the methods:


Using the formula

Example 1

Question

Solve the quadratic function ??x^{2} - 5x + 6 = 0?? for ??x?? by using the formula.
Show Solution


Completing The Square

Example 2

Question

Solve the quadratic function ??x^{2} - 5x + 6 = 0?? for ??x?? by completing the square.
Show Solution


Factorisation

Example 3

Question

Solve the quadratic function ??x^{2} - 5x + 6 = 0?? for ??x?? by using factorisation.
Show Solution


Exam Questions

The table below contains every exam question that has been asked on this topic, this includes normal papers from both January and June sittings, International papers and Specimen papers.

To see an exam question and solution simply click on the load question icon (), the question will appear below the table and the solution can be shown by clicking the "Show Solution" button that also appears.

The full exam paper and mark schemes are also available for download by clicking on the download icons in each row of the table ().

Load
Question
Exam Board Subject Paper Year Month Module Question
No.
Parts Total
Marks
Exam
Paper
Mark
Scheme
Edexcel Maths Standard 2005 June Core 1 3 n/a 6
Edexcel Maths Standard 2006 January Core 1 1 n/a 3
Edexcel Maths Standard 2006 January Core 1 9 (a) 2
Edexcel Maths Standard 2006 January Core 1 10 (a) 2
Edexcel Maths Standard 2006 June Core 1 9 (a), (b) 5
Edexcel Maths Standard 2008 June Core 1 2 n/a 3
Edexcel Maths Standard 2009 June Core 1 10 (a) 3
Edexcel Maths Standard 2010 January Core 1 9 (a) 3
Edexcel Maths Standard 2010 January Core 1 10 (a) 3
Edexcel Maths Standard 2010 June Core 1 4 (a) 2
Edexcel Maths Standard 2012 June Core 1 10 (a) 1
Edexcel Maths Standard 2013 January Core 1 1 n/a 3
Edexcel Maths Standard 2013 January Core 1 10 (a) 3


Exam Tips

  1. When attempting to find coefficients of an expression:
          Start by multiplying out any brackets. Take care when doing this with sign convention and correct terms.
          Rewriting into a standard form ??\color{red}{a}x^{2} + \color{green}{b}x + \color{blue}{c}?? makes comparisons easier.

  2. Remember, when factorising we are we are trying to rewrite the quadratic as the product of two linear expressions; ??\color{red}{a}x^{2} + \color{green}{b}x + \color{blue}{c} = (dx + e)(fx + g)??:
          Where the ??x^{2}?? term, ??\color{red}{a} = d \times f??
          Where the ??x?? term, ??\color{green}{b} = (d \times g) + (e \times f)??
          Where the constant term, ??\color{blue}{c} = e \times g??

  3. However in typical factorisation questions the factor ??\color{red}{a}?? of ??x^{2}?? is equal to 1 therefore factorised solution simplifies to: ??x^{2} + \color{green}{b}x + \color{blue}{c} = (x + e)(x + g)??
          Where the ??x?? term, ??\color{green}{b} = e + g??
          Where the constant term, ??\color{blue}{c} = e \times g??
          So in this case we are looking for two numbers that sum to give ??\color{green}{b}?? and multiply to give ??\color{blue}{c}??

  4. Factorisation takes practice to become quick at spotting the solution but look out for special cases like the 'difference of two squares'.
          When the equation takes the form ??(x^{2} - p^{2})?? this factorises into ??(x + p)(x - p)??.
         For example ??(x^{2} - 9) = (x^{2} - 3^{2}) = (x + 3)(x - 3)??
          When the equation takes the form ??(1 - p^{2}x^{2})?? this factorises into ??(1 + px)(1 - px)??.
         For example ??(1 - 9x^{2}) = (1 - 3^{2}x^{2}) = (1 + 3x)(1 - 3x)??

  5. You are less likely to make arithmetic errors using factorisation, but unfortunately some expressions can't be factorised. You can always use the formula, but be very careful substituting in your values for a, b and c, particularly with negative values.