Solve this by completing the square:
x² - 6x + 4 = (x - a)² + b
I don't get how to do this - I know how to solve equations by completing the square but it looks different to ones I have done. I'm thinking I need to re-arrange it but I'm not sure???
Please help!!!
Maths Help!! Solving equations by completing square?? ?
2008-12-31 4:28 pm
回答 (5)
2008-12-31 4:45 pm
✔ 最佳答案
I hope this helps:Add 5 to both sides
x^2 - 6x + 9 = (x - a)^2 + b + 5
Factor the left side:
(x - 3)^2 = (x - a)^2 + b + 5
From this we can pull values for a and b
a = 3
b + 5 = 0 -> b = -5
2009-01-01 1:19 am
x^2 - 6x + 4 = (x - a)^2 + b
x^2 - 3x - 3x + 4 = (x - a)^2 + b
x^2 - 3x - 3x + 9 + 4 - 9 = (x - a)^2 + b
[(x^2 - 3x) - (3x - 9)] - 5 = (x - a)^2 + b
[x(x - 3) - 3(x - 3)] - 5 = (x - a)^2 + b
(x - 3)(x - 3) - 5 = (x - a)^2 + b
(x - 3)^2 - 5 = (x - a)^2 + b
a = 3
b = -5
x^2 - 3x - 3x + 4 = (x - a)^2 + b
x^2 - 3x - 3x + 9 + 4 - 9 = (x - a)^2 + b
[(x^2 - 3x) - (3x - 9)] - 5 = (x - a)^2 + b
[x(x - 3) - 3(x - 3)] - 5 = (x - a)^2 + b
(x - 3)(x - 3) - 5 = (x - a)^2 + b
(x - 3)^2 - 5 = (x - a)^2 + b
a = 3
b = -5
2009-01-01 12:40 am
(x - 3)^2 - 5
so, a = 3 and b = -5
so, a = 3 and b = -5
2009-01-01 12:39 am
Equation: x^2-6x+4
(x^2-6/1 x)+4
(x^2-6/1 x+(6/2)^2)+4-1(6/2)^2
(x-6/2)^2+4-36/4
(x-3)^2-20/4
(x-3)^2-5
Compare this with (x-a)^2 +b
(x-3)^2-5 = (x - a)² + b
a=3
b=-5
(x^2-6/1 x)+4
(x^2-6/1 x+(6/2)^2)+4-1(6/2)^2
(x-6/2)^2+4-36/4
(x-3)^2-20/4
(x-3)^2-5
Compare this with (x-a)^2 +b
(x-3)^2-5 = (x - a)² + b
a=3
b=-5
2009-01-01 12:34 am
2x^2-6x+4+b
收錄日期: 2021-05-01 11:43:21
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https://hk.answers.yahoo.com/question/index?qid=20081231082852AA81k6c