題:
How to do it ? Please help me~~ (Maths)
Charles Leung
2006-10-18 17:57:47 UTC
10x^2+9x-22=0
二 答案:
?
2006-10-18 18:54:22 UTC
10x^2+9x-22=0 =

x^2 + 9/10 x - 22/10 = 0

x^2 +2 (x) (9 /20) + (9/20)^2 - (9/20)^2 - 22/10 = 0

(x + 9/20)^2 = (9/20)^2 + 22/10

(x + 9/20) ^2 = 81/400 + 22/10

(x + 9/20)^2 = 81/400 + 880/400

(x + 9/20)^2 = 961/400

(x + 9/20) = ± √ (961/400)

(x + 9/20) = ± (31/20)

x = ± (31/20) - 9/20

x = ± (22/20)

x = ± 11/10
Cynthia
2006-10-18 18:07:37 UTC
you can use the quadratic formula x=[-b+/-√(b^2-4ac)]/2a



now, a=10, b=9, c=-22

so,



x={-9+/-√[(9^2-4(10)(-22)]} / 2(10)

=[-9+/-√(81+880)] /20

=[-9+/-√(961)] /20

=(-9+/-31)/20

=(-9+31)/20 or =(-9-31)/20

=11/10 =2



2006-10-18 18:08:44 補充:

aiya....痴埋左tim...咁樣寫清楚d=(-9 31)/20 =11/10 or =(-9-31)/20=2



2006-10-18 18:10:17 補充:

="=........漏左個負tim.呢個岩啦!!!!!=(-9 31)/20 =11/10 or =(-9-31)/20=-2



2006-10-24 19:44:14 補充:

the answer is wrong!!!wrongwrongwrong!!!!

it should be....x=11/10 or x=-2

如果x=-11/10...咁就唔等啦...

10x^2+9x-22=-19.8

而唔係等於0...

yahoo知識有個唔好ge地方就係...choose左出黎ge answer未必係岩ge!!!


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