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How to solve quadratics by completing the square

How to solve quadratics by completing the square

Completing the square is one of the most versatile technique in quadratic equations and expressions. By completing the square we can solve a quadratic function to find the:

  • roots of the quadratic function
  • turning points (maximum or minimum point) of the quadratic function

Before we start solving quadratic functions let us look at what a square is. A square is a resultant of multiplying two identical numbers. For example, 9 is a square obtained from multiplying 3 and 3.

Knowing the properties of a square number will help us understand how completing the square works. This tutorial is not about memorising the method by about understanding how the method actually works.

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Squares of algebraic functions

Multiplying an algebraic function by itself will give us the square of the function. For example:

(x + 3)(x + 3) = (x + 3)2

We can also expand the brackets by multiplying every term in the first bracket by every term in the second bracket to get:

(x + 3)(x + 3) = x2 + 3x + 3x + 9

  • = x2 + 6x + 9

Now let us look at the relationship between the terms of the squared function and its corresponding expanded form.

(x + 3)2 = x2 + 6x + 9

  • we have 3 in the squared function → (x +3)2
  • on the expanded function we have 2×3 as the coefficient of x → x2 + 6x + 9
  • we have 32 as the last term → x2 + 6x + 9

This relationship applies to all quadratic perfect squares and let us prove it as follows:

(x + a)2 = (x + a)(x + a)

  • = x2 + ax + ax + a2
  • = x2 + 2ax + a2

This is very important because it shows us that for a square (x2 + 2ax + a2), the last term (a2) is actually the square of half the coefficient of x. In short, a quadratic expression is a square if when you take the coefficient of x, divide it by 2 and square it you get the last term. Now lets look at the following examples and see if they are squares:

  1. x2 + 10x + 36
    • the coefficient of x is 10
    • divide 10 by 2 to get 5
    • square the 5 to get 25
    • this means that the expression is not a perfect square and it would have been a perfect square if 36 had been replaced by 25.
  2. y2 + 8y + 16
    • the coefficient of y is 8
    • divide 8 by 2 to get 4
    • square the 4 to get 16
    • therefore the expression is a perfect square.

Completing the square

Completing the square is a process of converting a non square quadratic expression into a perfect square before we solve it.



Example 1: Complete the square and factorise the following quadratic expressions.

  1. x2 + x
  2. x2 + 2x
  3. x2 + 3x
  4. x2 + 4x
  5. x2 + 5x

Solution 1

If you have been following the tutorial very carefully you should now understand that we complete the square by adding the square of half the coefficient of x. This is what we are going to do and then we factorise the resultant quadratic function in the normal way.

  1. x2 + x + 0.52

    • = x2 + x + 0.25
    • = (x + 0.5)2
  2. x2 + 2x + 12

    • = x2 + 2x + 1
    • = (x + 1)2
  3. x2 + 3x + 1.52

    • = x2 + 3x + 2.25
    • = (x + 1.5)2
  4. x2 + 4x + 22

    • = x2 + 4x + 4
    • = (x + 2)2

After learning how to complete the square, let us now start solving the quadratic equations.

Example 2: Solve the following quadratic equations by completing the square.

  1. x2 + 8x + 5 = 0
  2. 2x2 + 8x + 5 = 0

Solution 2

  1. x^2 + 8x + 5 = 0

    • x^2 + 8x = -5

    • x^2 + 8x + 4^2= -5 + 4^2

    • x^2 + 8x + 4^2= 16-5

    • x^2 + 8x + 4^2= 11

    • (x + 4)^2= 11

    • \sqrt{(x + 4)^2}= \pm \sqrt{11}

    • x + 4= \pm \sqrt{11}

    • x = -4 \pm \sqrt{11}

    • x = -4 + \sqrt{11} \ or\  -4 - \sqrt{11}

    • x = -0.68 \ or\  -7.32 \ (to\ 2dp)

  2. 2x^2 + 8x + 5 = 0

    • 2x^2 + 8x = -5

    • \frac{2x^2}{2} + \frac{8x}{2} = \frac{-5}{2}

    • x^2 + 4x = \frac{-5}{2}

    • x^2 + 4x + 2^2= \frac{-5}{2} + 2^2

    • (x + 2)^2 = 4 - \frac{5}{2}

    • (x + 2)^2 = \frac{8 - 5}{2}

    • (x + 2)^2 = \frac{3}{2}

    • \sqrt{(x + 2)^2} = \pm \sqrt{\frac{3}{2}}

    • x + 2 = \pm \sqrt{\frac{3}{2}}

    • x =-2 \pm \sqrt{\frac{3}{2}}

    • x = -3.22 or -0.78\ (\ to \ 2dp)


Completing the square can also be used to derive the general quadratic formula.

Example 3: Derive the quadratic formula by completing the square

ax^2 + bx + c = 0

  • ax^2 + bx = -c

  • \frac{ax^2}{a} + \frac{bx}{a} = \frac{-c}{a}

  • x^2 + \frac{bx}{a} + (\frac{b}{2a})^2 = \frac{-c}{a} + (\frac{b}{2a})^2

  • (x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} + \frac{-c}{a}

  • (x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a}

  • (x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}

  • \sqrt{(x + \frac{b}{2a})^2} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}

  • x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}

  • x = - \frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}

  • x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

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