Completing the Square (Cambridge International IGCSE Further Maths)

Completing the square

What is completing the square?

• Completing the square is another way of writing a quadratic function

• It means rewriting $y=ax^{2}+bx+c$ in the form $y=a(x+p)^{2}+q$

  • The key point is that $x$ now only occurs once in the equation

• It can be used to solve quadratic equations, sketch their graphs and to find the coordinates of the turning point

How do I complete the square?

The method used will depend on the value of the coefficient of the $x^{2}$ term in $y=ax^{2}+bx+c$

• When $a=1$

  • $p$ is half of $b$
  • $q$ is $c-p^{2}$

• When a ≠ 1

  • First take a factor of $a$ out of the $x^{2}$ and $x$ terms
  • Then continue as above

Exam Hint

• Sometimes a question will explicitly use the phrase complete the square
• Sometimes a question will use the form $a ( x + p )^{2} + q$ without using the phrase completing the square

Worked Example Write $3x^{2}-12x+k$ in the form $a ( x + p )^{2} + k + q$, where $a,p$ and $q$ are constants to be found, and $k$ is an unknown constant.

The form required is 'completing the square' (do not be put off by the $k$, it is just a constant!).

STEP 1 - Take a factor of 3 out of the $x^{2}$ and $x$ terms - leaving $k$ avoids awkward fractions.

$3 x^{2} - 12 x + k = 3 ( x^{2} - 4 x ) + k$

STEPS 2 and 3 - Complete the square on the $( x^{2} - 4 x )$ part only. "$p=-\frac{4}{2}=-2$" and "$q = 0 - ( - 2 )^{2} = 4$".

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

STEP 4 - Expand and simplify.

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

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

i.e.$a=3,p=-2,q=-12$

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 called 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 = ±4 - 5
    • x = -1 or x = -9

• If the equation is ax² + bx + c = 0 with a number in front of x², then divide both sides by a first, before completing the square

Exam Hint

• When making x the subject to find the solutions at the end, don't expand the squared brackets back out again!

  • Remember to use ±√ to get two solutions

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

*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 * *

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

Square root both sides Include the ± sign to get two solutions * *

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

Add 2 to both sides * *

$x=\pm4+2$ * *

Work out each solution separately

Final answer: x = 6 or x = -2