COMPLETING THE SQUARE (PENYEMPURNAAN KUASA DUA)
III. And Do Some More
One important thing to remember: The coefficient (pekali) for x2 MUST be
1. For the example in Section II, it is already 1, so you don’t have to do anything.
But what happens when you get this?:
f(x) = 3x2 – 5x + 27
You have to ‘take out’ the 3. It’s a bit like factorising: (3x + 6) = 3(x + 2). If you have a number that cannot be divided by 3, divide it anyway, and leave a fraction inside.
3x2 – 5x + 27 | = (3 ÷3)x2 - (5 ÷ 3)x + (27 ÷ 3) |
= 3(x2 – 5/3x + 9) |
After you’ve done that, just deal with the expression (sebutan) inside the bracket as you did with the previous example. Remember, don’t throw away the 3!
f(x) | = 3x2 – 5x + 27 | |
= 3[x2 – 5/3x
+ 9] = 3[x2 – 5/3x + (5/6)2 - (5/6)2 + 9] |
<~(see iii) |
|
= 3[(x – 5/6)2
- (5/6)2< + 9] = 3[(x – 5/6)2 + 8 11/36] = 3(x – 5/6)2 + 8 11/12 |
<~(see iv) <~(see v) |
iii) b = 5/3, so you put (5/3 ÷ 2) in the brackets.
iv)
- (5/6)2
+ 9 = - 25/36 + 324/36
= 299/36
= 8 11/36
v) At the end, you must multiply the whole thing by 3 again. That's why I said not to throw away the 3!
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