Completing the Square

GCSELevel 6-7Level 8-9AQACambridge iGCSEEdexcelEdexcel iGCSEOCR

Completing the Square

Completing the square is a method of changing the way that a quadratic is expressed. There are two reasons we might want to do this, and they are

  1. To help us solve the quadratic equation.
  2. To find the coordinates of the minimum (or maximum) point of a quadratic graph.

Make sure you are happy with the following topics before continuing.

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Completing the Square Formula

What does “completing the square” mean? Well, it involves taking a quadratic equation, and expressing it in the form,

ax2+bx+c=a(x+d)2+etextcolor{black}{ax^2 + b x + c = a left(x + textcolor{red}{d} right)^2 + textcolor{blue}{e}}

where

d=b2a  textcolor{red}{d} textcolor{black}{=dfrac{b}{2a}} ,, and      e=c−b24a   ,,  textcolor{blue}{e} textcolor{black}{ =c-dfrac{b^2}{4a}}

or

e=c−ad2textcolor{blue}{e} textcolor{black}{=c-a}textcolor{red}{d}textcolor{black}{^2}

Level 6-7GCSEAQAEdexcelOCRCambridge iGCSEEdexcel iGCSE
Level 6-7GCSEAQAEdexcelOCRCambridge iGCSEEdexcel iGCSE

Skill 1: Completing the Square a=1a=1

Solving quadratics via completing the square can be tricky, first we need to write the quadratic in the form (x+d)2+e(x+textcolor{red}{d})^2 + textcolor{blue}{e} then we can solve it. Since a=1a=1, this can be done in 44 easy steps.

Example: By completing the square, solve the following quadratic x2+6x+3=1x^2+6x +3=1

Step 1: Rearrange the equation so it is =0=0

(−1)         x2+6x+3=1x2+6x+2=0begin{aligned}(-1),,,,,,,,,x^2+6x+3 &=1 x^2 +6x +2&=0end{aligned}

Step 2: Half the coefficient of xx, so in this case d=6÷2=3textcolor{red}{d}=6div 2=textcolor{red}{3}, and add it in the place of dtextcolor{red}{d}

(x+3)2+e(x+textcolor{red}{3})^2 + textcolor{blue}{e}

Step 3: Next we need to find etextcolor{blue}{e} which equals the constant at the end of the quadratic, +2+2, minus d2textcolor{red}{d^2}, then replace etextcolor{blue}{e} in the equation (e=c−d2textcolor{blue}{e} textcolor{black}{=c-}textcolor{red}{d}textcolor{black}{^2} as a=1textcolor{black}{a=1}).

e=2−9e= −7begin{aligned}textcolor{blue}{e} &= 2 -textcolor{maroon}{9} textcolor{blue}{e} &=  -7end{aligned}

(x+3)2 −7=0(x+textcolor{red}{3})^2  textcolor{blue}{-7} = 0

Step 4: Now we have the equation in this form we can solve the equation.

(+7)         (x+3)2−7=0()                    (x+3)2=7(−3)                           x+3=±7x=±7−3begin{aligned}(+7),,,,,,,,,(x+3)^2 -7 &= 0 (sqrt{}),,,,,,,,,,,,,,,,,,,,(x+3)^2 &= 7 (-3),,,,,,,,,,,,,,,,,,,,,,,,,,, x+3 &= pm sqrt{7} x &= pm sqrt{7}- 3end{aligned}

This gives the solutions to be

7−3sqrt{7} – 3 and −7−3-sqrt{7} – 3

Remember: A square root can have both a positive and negative solution

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Skill 2: Complete the square a>1a>1

When a≠1aneq1 things become a little trickier. The majority of the method is the same but with an additional factorisation step at the beginning.

Example: Write 3x2+5x−33x^2 + 5x-3 in the form a(x+d)2+etextcolor{limegreen}{a}(x+textcolor{red}{d})^2+textcolor{blue}{e}

Step 1: Factorise the first two terms by the coefficient in front of x2x^2, this now becomes atextcolor{limegreen}{a}

3(x2+53x)−3textcolor{limegreen}{3}bigg(x^2 + dfrac{5}{3}x bigg)-3

Step 2: Half the coefficient of xx and write it in the place of dtextcolor{red}{d}

53÷2=56dfrac{5}{3} div 2 = textcolor{red}{dfrac{5}{6}}

3(x+56)2+etextcolor{limegreen}{3}bigg(x+textcolor{red}{dfrac{5}{6}}bigg)^2+textcolor{blue}{e}

Step 3: Next we need to find etextcolor{blue}{e} which equals the constant at the end of the quadratic, −3-3, minus the ‘non-xx‘ result from expanding the brackets. (e=c−ad2textcolor{blue}{e} textcolor{black}{=c-a}textcolor{red}{d}textcolor{black}{^2})

3(x+56)2=3(x+56)(x+56)=3x2+5x+25123bigg(x+dfrac{5}{6}bigg)^2 = 3bigg(x+dfrac{5}{6}bigg)bigg(x + dfrac{5}{6}bigg) = 3x^2 +5x + dfrac{25}{12}

3(x+56)2−3−25123bigg(x+dfrac{5}{6}bigg)^2 – 3 – dfrac{25}{12}

=3(x+56)2−6112= 3bigg(x+dfrac{5}{6}bigg)^2 -dfrac{61}{12}

So the completed square is

3(x+56)2−6112textcolor{limegreen}{3}bigg(x+textcolor{red}{dfrac{5}{6}}bigg)^2 textcolor{blue}{-dfrac{61}{12}}

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Completing the Square Example Questions

Question 1: Write m2+5m+6m^2 + 5m + 6 in the form (x+a)2+b(x+a)^2+b, where aa and bb are constants to be determined.

[3 marks]

Level 6-7GCSE AQAEdexcelOCRCambridge iGCSEEdexcel iGCSE

The coefficient of mm term is 55, and half of 55 is 52frac{5}{2}, so we get

 

m2+5m+6=(m+52)2+6−(52)2m^2+5m+6=left(m+dfrac{5}{2}right)^2 + 6 – left(dfrac{5}{2}right)^2

 

Considering the value of bb,

 

6−(52)2=6−254=244−254=−146-left(dfrac{5}{2}right)^2 = 6 – dfrac{25}{4}=dfrac{24}{4}-dfrac{25}{4}=-dfrac{1}{4}

 

Now we know the constant that goes outside the bracket, the final result of completing the square is,

 

(m+52)2−14left(m+dfrac{5}{2}right)^2-dfrac{1}{4}

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Question 2: Write 2x2−8x+102x^2 – 8x + 10 in the form a(x+b)2+ca(x + b)^2 + c, where a,b,a, b, and cc are constants to be determined.

[4 marks]

Level 8-9GCSE AQAEdexcelOCRCambridge iGCSEEdexcel iGCSE

In order to be able to apply our normal process of completing the square, we need to take a factor of 22 out of this whole expression:

 

2x2−8x+10=2(x2−4x+5)2x^2 – 8x + 10 = 2(x^2 – 4x + 5)

 

Now it looks more familiar, the coefficient of the xx term is −4-4, half of which is −2-2, so we get:

 

2(x2−4x+5)=2[(x−2)2+5−(−2)2]=2[(x−2)2+1]2(x^2 – 4x + 5) = 2left[(x – 2)^2 + 5 – (-2)^2right] = 2[(x – 2)^2 + 1]

 

Multiplying through by 22 we find,

 

2[(x−2)2+1]=2(x−2)2+22left[(x -2)^2 + 1right] = 2(x – 2)^2 + 2

 

which is in the form asked for in the question.

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Question 3: Write x2−2mx+nx^2 – 2mx + n in the form a(x+b)2+ca(x + b)^2 + c, where a,b,a, b, and cc are constants to be determined.

[3 marks]

Level 6-7GCSE AQAEdexcelOCRCambridge iGCSEEdexcel iGCSE

The coefficient of xx term is −2m-2m, and half of −2m-2m is −m-m, so we get

 

x2−2mx+n=(x−m)2+n−(−m)2x^2-2mx+n=(x-m)^2 + n – (-m)^2

 

This simplifies so that the final result of completing the square is,

 

(x−m)2+(n−m2)(x-m)^2 + (n – m^2)

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Question 4: Use completing the square to find the exact solutions of, z2+14z−1=0z^2 + 14z – 1 = 0

[3 marks]

Level 6-7GCSE AQAEdexcelOCRCambridge iGCSEEdexcel iGCSE

The coefficient of the zz term is 1414, half of which is 77, so we get:

 

z2+14z−1=(z+7)2−1−72=(z+7)2−50z^2 + 14z – 1 = (z + 7)^2 – 1 – 7^2 = (z + 7)^2 – 50

 

Meaning our equation is now

 

(z+7)2−50=0(z + 7)^2 – 50 = 0

 

Now we must rearrange this equation to make zz the subject,

 

(z+7)2=50z+7=±50z=−7±50begin{aligned} (z + 7)^2 &= 50 z + 7 & = pm sqrt{50} z &= -7 pm sqrt{50}end{aligned}

 

So, the two solutions are,

x=−7+50x = -7 + sqrt{50} and x=−7−50x = -7 – sqrt{50}

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Question 5: Use completing the square to find the exact solutions of, 4x=3−x24x= 3-x^2

[5 marks]

Level 8-9GCSE AQAEdexcelOCRCambridge iGCSEEdexcel iGCSE

We have to start by writing the equation in a more familiar form,

 

 3−4x−x2=03-4x-x^2=0

 

In this case, we have a=−1a=-1 so now completing the square we get,

 

 3−4x−x2=−(x+2)2+3+43-4x-x^2 = -(x+2)^2 +3+4

 

Now to solve this quadratic we must rearrange it to make xx the subject,

 

−(x+2)2+7=0−(x+2)2=−7(x+2)2=7x+2=±7x=−2±7begin{aligned} -(x+2)^2 +7 &= 0 -(x+2)^2 &= -7 (x+2)^2 &= 7 x+2 & = pm sqrt{7} x&= -2 pm sqrt{7}end{aligned}

 

So, the two solutions are,

x=−2+7x = -2 + sqrt{7} and x=−2−7x = -2 – sqrt{7}

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Specification Points Covered

Algebra – 18. solve quadratic equations (including those that require rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph

Completing the Square Worksheet and Example Questions

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(NEW) Completing the Square (a=1) Exam Style Questions - MME

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(NEW) Completing the Square (a>1) Exam Style Questions - MME

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Completing the Square Drill Questions

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Completing the Square - Drill Questions

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Completing the Square Hard - Drill Questions

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Related Topics

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Powers and Roots

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Expanding Brackets

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Surds

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Rearranging Formulae

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