1.Use the factor method to solve the following quadratic equation in one
unknown.
(a) 8y-y^2=16
(b) (z+1)(z+9)=4(z+1)
2. Solve the following quadratic equation using the factor method.(呢一題我唔知點解要計，可可用中文解一次比我聽啊！！ｔｈｘ
(u-1)(u+3)=u+3
(u-1)(u+3)-(u+3)=0
(u-1-1)(u+3)=0 (<------我唔明呢一個steps)
(u-2)(u+3)=0
maths promble(quadratic equation)
2008-12-10 4:49 am
回答 (3)
2008-12-10 5:08 am
✔ 最佳答案
1. (a). ...........8y-y^2=16.........-y^2 + 8y - 16 = 0
..........y^2 - 8y + 16 = 0
...............( y - 4 )^2 = 0
.......................y - 4 = 0
............................y = 4 ( 重根 )
1. (b)........(z+1)(z+9) = 4(z+1)
........z^2 + z + 9z + 9 = 4z + 4
..............z^2 + 6z + 5 = 0
............(z + 5)(z + 1) = 0
......................z = -5 或 -1
2. ........(u-1)(u+3) = u+3
...(u-1)(u+3)-(u+3) = 0
.......(u+3)[(u-1)-1] = 0 [ 因為u + 3在這題中是共同因子, 所以可抽出來 ]
............(u+3)(u-2) = 0
..............u = -3 或 2
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2008-12-10 5:01 am
1a.
8y-y^2=16
y^2-8y+16=0
y^2-2(y)(4)+4^2=0
[y-4]^2=0
y=4 (repeated)
1b.
(z+1)(z+9)=4(z+1)
(z+1)(z+9)-4(z+1)=0
(z+1)[(z+9)-4]=0
(z+1)(z+5)=0
z=-1 or z=-5
2.
(u-1)(u+3)=u+3
(u-1)(u+3)-1(u+3)=0
(u+3)[(u-1)-1]=0.................抽(u+3)做factor，「ab-b=(a-1)b」
(u+3)(u-2)=0
u=-3 or u=2
8y-y^2=16
y^2-8y+16=0
y^2-2(y)(4)+4^2=0
[y-4]^2=0
y=4 (repeated)
1b.
(z+1)(z+9)=4(z+1)
(z+1)(z+9)-4(z+1)=0
(z+1)[(z+9)-4]=0
(z+1)(z+5)=0
z=-1 or z=-5
2.
(u-1)(u+3)=u+3
(u-1)(u+3)-1(u+3)=0
(u+3)[(u-1)-1]=0.................抽(u+3)做factor，「ab-b=(a-1)b」
(u+3)(u-2)=0
u=-3 or u=2
參考: ME
2008-12-10 4:58 am
1)a)8y-y^2=16
y^2-8y+16=0
(y-4)^2=0
y=4
b)(z+1)(z+9)=4(z+1)
(z+1)(z+9)-4(z+1)=0
(z+1)(z+9-4)=0-->抽 common factor(z+1)
(z+1)(z+5)=0
z=-1 or z=-5
2)你唔明的step 都係抽common factor (u+3)
即:AB+BC=0
(A+C)xB=0
A=u-1
C=-1
B=u+3
y^2-8y+16=0
(y-4)^2=0
y=4
b)(z+1)(z+9)=4(z+1)
(z+1)(z+9)-4(z+1)=0
(z+1)(z+9-4)=0-->抽 common factor(z+1)
(z+1)(z+5)=0
z=-1 or z=-5
2)你唔明的step 都係抽common factor (u+3)
即:AB+BC=0
(A+C)xB=0
A=u-1
C=-1
B=u+3
收錄日期: 2021-04-29 21:42:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081209000051KK01487