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標題:

How to do it ? Please help me~~ (Maths)

發問:

10x^2+9x-22=0

最佳解答:

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

其他解答:

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!!!6A560DB67A298191