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Solution by completing the square

We have seen that expressions of the form:

a 2 x 2 - b 2

are known as differences of squares and can be factorised as follows:

( a x - b ) ( a x + b ) .

This simple factorisation leads to another technique to solve quadratic equations known as completing the square .

We demonstrate with a simple example, by trying to solve for x in:

x 2 - 2 x - 1 = 0 .

We cannot easily find factors of this term, but the first two terms look similar to the first two terms of the perfect square:

( x - 1 ) 2 = x 2 - 2 x + 1 .

However, we can cheat and create a perfect square by adding 2 to both sides of the equation in [link] as:

x 2 - 2 x - 1 = 0 x 2 - 2 x - 1 + 2 = 0 + 2 x 2 - 2 x + 1 = 2 ( x - 1 ) 2 = 2 ( x - 1 ) 2 - 2 = 0

Now we know that:

2 = ( 2 ) 2

which means that:

( x - 1 ) 2 - 2

is a difference of squares. Therefore we can write:

( x - 1 ) 2 - 2 = [ ( x - 1 ) - 2 ] [ ( x - 1 ) + 2 ] = 0 .

The solution to x 2 - 2 x - 1 = 0 is then:

( x - 1 ) - 2 = 0

or

( x - 1 ) + 2 = 0 .

This means x = 1 + 2 or x = 1 - 2 . This example demonstrates the use of completing the square to solve a quadratic equation.

Method: solving quadratic equations by completing the square

  1. Write the equation in the form a x 2 + b x + c = 0 . e.g. x 2 + 2 x - 3 = 0
  2. Take the constant over to the right hand side of the equation. e.g. x 2 + 2 x = 3
  3. If necessary, make the coefficient of the x 2 term = 1, by dividing through by the existing coefficient.
  4. Take half the coefficient of the x term, square it and add it to both sides of the equation. e.g. in x 2 + 2 x = 3 , half of the coefficient of the x term is 1 and 1 2 = 1 . Therefore we add 1 to both sides to get: x 2 + 2 x + 1 = 3 + 1 .
  5. Write the left hand side as a perfect square: ( x + 1 ) 2 - 4 = 0
  6. You should then be able to factorise the equation in terms of difference of squares and then solve for x : ( x + 1 - 2 ) ( x + 1 + 2 ) = 0

Solve:

x 2 - 10 x - 11 = 0

by completing the square

  1. x 2 - 10 x - 11 = 0
  2. x 2 - 10 x = 11
  3. The coefficient of the x 2 term is 1.

  4. The coefficient of the x term is -10. Therefore, half of the coefficient of the x term will be ( - 10 ) 2 = - 5 and the square of it will be ( - 5 ) 2 = 25 . Therefore:

    x 2 - 10 x + 25 = 11 + 25
  5. ( x - 5 ) 2 - 36 = 0
  6. ( x - 5 ) 2 - 36 = 0
    [ ( x - 5 ) + 6 ] [ ( x - 5 ) - 6 ] = 0
  7. [ x + 1 ] [ x - 11 ] = 0 x = - 1 or x = 11

Solve:

2 x 2 - 8 x - 16 = 0

by completing the square

  1. 2 x 2 - 8 x - 16 = 0
  2. 2 x 2 - 8 x = 16
  3. The coefficient of the x 2 term is 2. Therefore, divide both sides by 2:

    x 2 - 4 x = 8
  4. The coefficient of the x term is -4; ( - 4 ) 2 = - 2 and ( - 2 ) 2 = 4 . Therefore:

    x 2 - 4 x + 4 = 8 + 4
  5. ( x - 2 ) 2 - 12 = 0
  6. [ ( x - 2 ) + 12 ] [ ( x - 2 ) - 12 ] = 0
  7. [ x - 2 + 12 ] [ x - 2 - 12 ] = 0 x = 2 - 12 or x = 2 + 12
  8. Leave the left hand side written as a perfect square

    ( x - 2 ) 2 = 12
  9. x - 2 = ± 12
  10. Therefore x = 2 - 12    or     x = 2 + 12

    Compare to answer in step 7.

Khan academy video on solving quadratics - 1

Solution by completing the square

Solve the following equations by completing the square:

  1. x 2 + 10 x - 2 = 0
  2. x 2 + 4 x + 3 = 0
  3. x 2 + 8 x - 5 = 0
  4. 2 x 2 + 12 x + 4 = 0
  5. x 2 + 5 x + 9 = 0
  6. x 2 + 16 x + 10 = 0
  7. 3 x 2 + 6 x - 2 = 0
  8. z 2 + 8 z - 6 = 0
  9. 2 z 2 - 11 z = 0
  10. 5 + 4 z - z 2 = 0

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Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
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