Completing the Square (Cambridge (CIE) IGCSE Maths: Extended): Revision Note

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 + c 

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

$x^2+bx$ is the same as ${\left(x+p\right)}^2-p^2$ where $p$ is half of $b$

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

  • Is $x^2+2x$ the same as ${\left(x+1\right)}^2-1^2$?
    

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

• This works for negative values of b too

  •  $x^2-20x$ can be written as ${\left(x-10\right)}^2-{\left(-10\right)}^2$ which is ${\left(x-10\right)}^2-100$

  • 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

  • $a{x}^2+bx+c=a\left[x^2+\frac{b}{a}x\right]+c$

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

• Then complete the square on the bit inside the brackets: $x^2+\frac{b}{a}x$

  • This gives $a\left[{\left(x+p\right)}^2-p^2\right]+c$

    • where p is half of $\frac{b}{a}$

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

  • $a{\left(x+p\right)}^2-a{p}^2+c$

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

  • Usually you are asked to give your final answer in the form  $a{\left(x+p\right)}^2+q$ 

  • 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 = $\frac{4}{2}$ = 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 $-x^2+bx+c$, 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 $y={\left(x+p\right)}^2+q$ then the turning point is at $\left(-p,q\right)$

    • Notice the negative sign in the x-coordinate

    • This links to transformations of graphs

    • A translation of $y=x^2$ by p to the left and q up

  • If $y=a{\left(x+p\right)}^2+q$ then the turning point is still at $\left(-p,q\right)$

    • 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

• 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

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

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)

    • $a{x}^2+bx+c=a\left[x^2+\frac{b}{a}x\right]+c$ and so on

How does completing the square link to the quadratic formula?

• 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,  $x=\frac{-10\pm \sqrt{64}}{2}=-5\pm \frac{8}{2}$ = -5 ± 4 (the same structure)