2條F.4數學問題 高手幫幫忙!

2008-11-09 1:48 am
1. solve the equation (x/2+2)/3=5

2.suppose the graph of y=2x^2+x-1 is given . Find the equation of the straight lines needed to be drawn on the given graph so as to solve the equation.
4x^2-3x-5=0


詳細解答pls

回答 (2)

2008-11-09 1:58 am
✔ 最佳答案
1. (x/2 + 2) /3 = 5
x/2 + 2 = 15
x/2 = 15 -2 = 13
x = 13 x 2 = 26.
2.
4x^2 - 3x - 5 = 0
2(2x^2 + x - 1) - 5x - 3 = 0
2x^2 + x - 1 = (5x + 3)/2
so the required straight line is y = (5x+3)/2.
2008-11-09 2:33 am
1.x/2 2=15
x/2=13
x=26

cuz equation 2 :4x^2-3x-5=0
cuz x=(-b (b^2-4ac)^ 1/2)/2a or x=(-b-(b^2-4ac)^ 1/2)/2a
x1,2=(3 -(89^1/2))/8
puts into y=2x^2 x-1
when x = (3 89 ^ 1/2) / 8 , y = (39 5*(89)^1/2)/16
when x = (3 - 89 ^ 1/2) / 8 , y = (39 - 4*(89)^1/2)/16
related these two points to draw the straight line....

2008-11-08 18:36:28 補充:
oh~上面戈位第二題岩d....甘拜下風

2008-11-08 18:41:11 補充:
更正...不能顯示
x1,2=(3 +-(89^1/2))/8
puts into y=2x^2 x-1
when x = (3+ 89 ^ 1/2) / 8 , y = (39 +5*(89)^1/2)/16

但係個答案都差好遠...點解....(I'm a F5 student in macau....I learn maths in chinese...so I'm guessing the meaning of question 2....)
參考: myself....


收錄日期: 2021-04-25 22:39:22
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