Completing the Square (Cambridge International IGCSE Mathematics)

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 $( x + p )^{2} - p^{2}$ where $p$ is half of $b$

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

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

    • 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 $( x - 10 )^{2} - ( - 10 )^{2}$ which is $( x - 10 )^{2} - 100$
  • 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 + 10*x, *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

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

    • 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 $a [ ( x + p )^{2} - p^{2} ] + c$

    • 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

  • $a ( x + p )^{2} - a p^{2} + c$

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

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

  • 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 $y = ( x + p )^{2} + q$ then the turning point is at $( - p , q )$

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

    • 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

Exam Hint

• 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 $y=x^{2}+6x-11$.

Answer:

Find half of +6 (call this p)

$p=\frac{6}{2}=3$

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

$x^{2}+6x$ is the same as $( x + 3 )^{2} - 3^{2}$

Put this result into the equation of the curve

$y = ( x + 3 )^{2} - 3^{2} - 11$

Simplify the numbers

$y = ( x + 3 )^{2} - 20$

Use the fact that the turning point of $y = ( x + p )^{2} + q$ is at $( - p , q )$* * Here p = 3 and q= -20

Final answer: turning point at (-3, -20)

(b) Write $-3x^{2}+12x+24$ in the form $a ( x + p )^{2} + q$.

Answer:

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

$- 3 [ x^{2} - 4 x ] + 24$

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

$- 3 [ ( x - 2 )^{2} - ( - 2 )^{2} ] + 24$

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

$- 3 [ ( x - 2 )^{2} - 4 ] + 24$

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

$- 3 ( x - 2 )^{2} + 12 + 24$

Simplify the numbers

$- 3 ( x - 2 )^{2} + 36$

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

$-3(x-2)^{2}+36$

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 + 10*x *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

Exam Hint

• 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)

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

Worked Example Solve $2x^{2}-8x-24=0$ by completing the square.

Answer:

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

$x^{2}-4x-12=0$

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

$( x - 2 )^{2} - ( - 2 )^{2} - 12 = 0$

Simplify the numbers

$\begin{array}{rcl} ( x - 2 )^{2} - 4 - 12 & = 0 \\ ( x - 2 )^{2} - 16 & = 0 \end{array}$

Add 16 to both sides

$( x - 2 )^{2} = 16$

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

$x-2=\pm\sqrt{16}=\pm4$

Add 2 to both sides

$x=2\pm4$

Work out each solution separately

Final answer: x = 6 or x = -2