Completing the Square (Cambridge (CIE) IGCSE Maths)

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Completing the Square

How can I rewrite the first two terms of a quadratic expression as the difference of two squares?

  • Look at the quadratic expression x2 + bx +

  • The first two terms can be written as the difference of two squares using the following rule

x squared plus b x is the same as open parentheses x plus p close parentheses squared minus p squared where p is half of b

  • Check this is true by expanding the right-hand side

    • Is x squared plus 2 x the same as open parentheses x plus 1 close parentheses squared minus 1 squared?

      • Yes: (x + 1)(x + 1) - 12 = x2 + 2x + 1 - 1 = x2 + 2x

  • This works for negative values of b too

    •  x squared minus 20 x can be written as open parentheses x minus 10 close parentheses squared minus open parentheses negative 10 close parentheses squared which is open parentheses x minus 10 close parentheses squared minus 100

    • A negative b does not change the sign at the end

How do I complete the square?

  • Completing the square is a way to rewrite a quadratic expression in a form containing a squared bracket

  • To complete the square on x2 + 10x + 9

    • Use the rule above to replace the first two terms, x2 + 10x, with (x + 5)2 - 52

    • then add 9:  (x + 5)2 - 52 + 9

    • simplify the numbers:  (x + 5)2 - 25 + 9

    • answer: (x + 5)2 - 16 

How do I complete the square when there is a coefficient in front of the x2 term?

  • You first need to take a out as a factor of the x2 and x terms only

    • Factorise the first two terms

    • a x squared plus b x plus c equals a open square brackets x squared plus b over a x close square brackets plus c

      • Use square-shaped brackets here to avoid confusion with round brackets later

  • Then complete the square on the bit inside the brackets: x squared plus b over a x

    • This gives a open square brackets open parentheses x plus p close parentheses squared minus p squared close square brackets plus c

      • where p is half of b over a

  • Finally multiply this expression through by a (from outside the square brackets) and add the c on to the end

    • a open parentheses x plus p close parentheses squared minus a p squared plus c

      • This looks far more complicated than it is in practice!

    • Usually you are asked to give your final answer in the form  a open parentheses x plus p close parentheses squared plus q 

    • For example, y = 4x2 + 16x + 5

      • Factorise out 'a' on the right-hand side (use square brackets)

        • y = 4[x2 + 4x] + 5

      • Replace x2 + 4x with (x + 2)2 - 22  (because p = 4 over 2 = 2)

        • y = 4[(x + 2)2 - 22] + 5

      • Simplify the terms inside the square brackets

        • y = 4[(x + 2)2 - 4] + 5

      • Multiply everything inside the square brackets by 4

        • y = 4(x + 2)2 - 16 + 5

      • Simplify to get the final answer

        • y = 4(x + 2)2 - 11

  • For quadratics like negative x squared plus b x plus c, do the above but with a = -1

How do I find the turning point by completing the square?

  • Completing the square helps us find the turning point on a quadratic graph

    • If y equals open parentheses x plus p close parentheses squared plus q then the turning point is at open parentheses negative p comma q close parentheses

      • Notice the negative sign in the x-coordinate

      • This links to transformations of graphs

      • A translation of y equals x squared by p to the left and q up

    • If y equals a open parentheses x plus p close parentheses squared plus q then the turning point is still at open parentheses negative p comma q close parentheses

      • The a does not change the coordinates

      • The turning point is a minimum point if a > 0

      • or a maximum point if a < 0

  • This can also help you create the equation of a quadratic when given the turning point

Completing the square Notes Diagram 3, A Level & AS Level Pure Maths Revision Notes
  • It can also be used to prove or show results using the fact that any squared term, such as the squared bracket (x ± p)2, will always be greater than or equal to 0

    • You cannot square a number and get a negative value

    • The smallest a squared term can be is 0

Completing the square Notes Diagram 4, A Level & AS Level Pure Maths Revision Notes

Examiner Tips and Tricks

  • To know if you have completed the square correctly, expand your answer to check

Worked Example

(a) By completing the square, find the coordinates of the turning point on the graph of y equals x squared plus 6 x minus 11.

Find half of +6 (call this p)

p equals 6 over 2 equals 3

Write x2 + 6x in the form (x + p)2 - p2

x squared plus 6 x is the same as open parentheses x plus 3 close parentheses squared minus 3 squared

Put this result into the equation of the curve

y equals open parentheses x plus 3 close parentheses squared minus 3 squared minus 11 

Simplify the numbers

y equals open parentheses x plus 3 close parentheses squared minus 20 

Use the fact that the turning point of y equals open parentheses x plus p close parentheses squared plus q is at open parentheses negative p comma q close parentheses 
Here p = 3 and q = -20

turning point at (-3, -20)

(b) Write negative 3 x squared plus 12 x plus 24 in the form a open parentheses x plus p close parentheses squared plus q

Factorise -3 out of the first two terms only
Use square-shaped brackets

negative 3 open square brackets x squared minus 4 x close square brackets plus 24 

Complete the square on the x2 - 4x inside the brackets
Write in the form (x + p)2 - p2 where p is half of -4

negative 3 open square brackets open parentheses x minus 2 close parentheses squared minus open parentheses negative 2 close parentheses squared close square brackets plus 24 

Simplify the numbers inside the brackets
(-2)2 is 4

negative 3 open square brackets open parentheses x minus 2 close parentheses squared minus 4 close square brackets plus 24 

Multiply -3 by all the terms inside the square brackets
(You do not multiply -3 by the 24)

negative 3 open parentheses x minus 2 close parentheses squared plus 12 plus 24 

Simplify the numbers

negative 3 open parentheses x minus 2 close parentheses squared plus 36 

This is now in the form a(x + p)2 + q where a = -3, p = -2 and q = 36

negative 3 stretchy left parenthesis x minus 2 stretchy right parenthesis squared plus 36

Solving by Completing the Square

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

  • To solve x2 + bx + c = 0 

    • replace the first two terms, x2 + bx, with (x + p)2 - p2 where p is half of b

    • This is 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 + 5 = ±4

      • x  = -5 ±4

      • x  = -1 or x  = -9

  • It also works with numbers that lead to surds

    • The answers found will be in exact (surd) form

Examiner Tips and Tricks

  • When making x the subject to find the solutions, don't expand the squared bracket back out again!

    •  Remember to use ±√ to get two solutions

How do I solve by completing the square when there is a coefficient in front of the x2 term?

  • If the equation is ax2 + bx + c = 0 with a number (other than 1) in front of x2

    • you can divide both sides by a first (before completing the square)

      • For example 3x2 + 12x + 9 = 0

      • Divide both sides by 3

        • x2 + 4x + 3 = 0

      • Complete the square on this easier equation

  • This trick only works when completing the square to solve a quadratic equation

    • i.e. it has an "=0" on the right-hand side

  • Don't do this when using completing the square to rewrite a quadratic expression in a new form

    • i.e. when there is no "=0"

    • For that, you must factorise out the a (but not divide by it)

      • a x squared plus b x plus c equals a open square brackets x squared plus b over a x close square brackets plus c and so on

  • The quadratic formula actually comes from completing the square to solve ax2 + bx + c = 0

    • a, b and c are left as letters when completing the square

      • This makes it 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 so x  = -5 ± 4 (as above) 

      • by the quadratic formula,  x equals fraction numerator negative 10 plus-or-minus square root of 64 over denominator 2 end fraction equals negative 5 plus-or-minus 8 over 2 = -5 ± 4 (the same structure)

Worked Example

Solve 2 x squared minus 8 x minus 24 equals 0 by completing the square.

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

x squared minus 4 x minus 12 equals 0 

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

open parentheses x minus 2 close parentheses squared minus open parentheses negative 2 close parentheses squared minus 12 equals 0 

Simplify the numbers

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x minus 2 close parentheses squared minus 4 minus 12 end cell equals 0 row cell open parentheses x minus 2 close parentheses squared minus 16 end cell equals 0 end table 

Add 16 to both sides

open parentheses x minus 2 close parentheses squared equals 16

Take the square root of both sides
Include the ± sign to get two solutions

x minus 2 equals plus-or-minus square root of 16 equals plus-or-minus 4 

Add 2 to both sides

x equals 2 plus-or-minus 4

Work out each solution separately

x = 6  or  x = -2

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Mark Curtis

Author: Mark Curtis

Expertise: Maths

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

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Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.