COMPLETING THE SQUARE (PENYEMPURNAAN KUASA DUA)

II. How to Do It

Of course, you won’t get it so simple in tests… in Section I you could do x2 + 6x + 8 because I already told you that x2 + 6x + 9 = (x + 3)2. Now look at this:

     x2 - 5x + 27

 

Jeng jeng jeng! You have no idea what to turn 27 into, right? Don’t worry, there’s a systematic way to do it. Look at it as ax2 + bx + c.

  1. Prepare a space between bx and c
    x2 - 5x +                       + 27
  1. Take HALF of b and square it. Here, b is 5, so you get (5/2)2
  1. Add that value into the space in your equation, and take it away again (buruk siku mah)
    x2 - 5x + (5/2)2 - (5/2)2 + 27

This is because [(5/2)- (5/2)2] equals nothing. So even though it LOOKS like you added something to the equation, you actually didn’t (you buruk siku: give people something and take it away again).

 

  1. Look at it like this:
    x2 - 5x + (5/2)2 - (5/2)2 + 25
  1. The bold part can be factorised. Look at the ‘squares’: x2 and (5/2)2. Remove the square (kuasa dua) signs from them, and get rid of the middle part (5x) totally. You now have (x - 5/2).

    Notice the highlighted operator (tanda): ‘-’. You must copy the operator of b.
  1. Put a square on (x - 5/2), and put in the rest of the equation
    (x - 5/2) 2 - (5/2)2 + 25
  1. Simplify the other part
    - (5/2)2 + 25   = - 25/4 + 25
                           = 75/4
  1. Final step! Put them together
    (x - 5/2) 2 + 75/4

To recap,

 

f(x)     = x2 – 5x + 27

  = x2 - 5x +      &nnbsp;             + 27

  = x2 - 5x + (5/2)2 - (5/2)2 + 27    <~(see i)

  = (x - 5/2) 2 - (5/2)2 + 27

  = (x - 5/2) 2 + 83/4                      <~(see ii)

 

i)     b is 5, so you put (5 4 2) in the brackets.

ii)     - (5/2)2 + 27 = - 25/4 + 108/4
            = 83/4

The red part is just to show you what’s going on. Your teacher won’t be happy if you wrote the red part in your paper.

Page: 1 | 2 | 3 | 4
The Notes Portal Home

1