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- Factorising and completing the square
Factorising and completing the square
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Completing the square rewrites a quadratic so you can read off its turning point and solve it exactly.
Factorising
For ax2 + bx + c, find two numbers that multiply to ac and add to b, then split the middle term.
3x2 − 10x − 8: ac = −24, use −12 and 2
3x2 − 12x + 2x − 8 = 3x(x − 4) + 2(x − 4) = (3x + 2)(x − 4)
3x2 − 10x − 8: ac = −24, use −12 and 2
3x2 − 12x + 2x − 8 = 3x(x − 4) + 2(x − 4) = (3x + 2)(x − 4)
Completing the square
x2 + bx + c = (x + b2)2 − (b2)2 + c
x2 + 8x + 3 = (x + 4)2 − 16 + 3 = (x + 4)2 − 13
The minimum value is −13, when x = −4.
x2 + 8x + 3 = (x + 4)2 − 16 + 3 = (x + 4)2 − 13
The minimum value is −13, when x = −4.
When a is not 1
Take a out of the x terms first.
2x2 + 12x + 5 = 2(x2 + 6x) + 5
= 2[(x + 3)2 − 9] + 5 = 2(x + 3)2 − 13
2x2 + 12x + 5 = 2(x2 + 6x) + 5
= 2[(x + 3)2 − 9] + 5 = 2(x + 3)2 − 13
Write 5 + 4x − x2 in the form a − (x − b)2 and state its maximum value.
- 5 + 4x − x2 = 5 − (x2 − 4x)
- x2 − 4x = (x − 2)2 − 4
- 5 − [(x − 2)2 − 4] = 9 − (x − 2)2
Answer: 9 − (x − 2)2, maximum value 9 when x = 2
x2 + bx + c = (x + b2)2 − (b2)2 + c. Take out a first if a ≠ 1. (x + p)2 + q has its minimum value q when x = −p.
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