Solve the quadratic equation by using this formula,or any other method.
-8(2x+1)(2x+1)=-64
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我想知係點計呀><"
點解我會計左開方128既!?
Quadratic equation!!!!!
2007-09-14 6:13 am
回答 (4)
2007-09-14 6:19 am
✔ 最佳答案
(2X+1)^2=82X+1=正負開方8
2x=-1正負開方8
x=-1正負開方8/2
2007-09-14 7:22 am
method 1:
- 8 (2x+1) (2x+1) = - 64
(2x+1)^2 = 8
2x + 1 = ± √8
2x = - 1 ± √8
x = ( - 1 ± √8 ) / 2
method 2:
-8(2x+1)(2x+1)=-64
(2x+1)^2=8
4 x^2 + 4x + 1 = 8
4 x^2 + 4x - 7 = 0
x = { - 4 ± √[4^2 - 4*4*(-7)] } / (2*4)
x = ( - 4 ± √128 ) / 8
x = ( - 4 ± 8 √2 ) / 8
x = ( - 1 ± 2 √2 ) / 2
x = ( - 1 ± √8 ) / 2
- 8 (2x+1) (2x+1) = - 64
(2x+1)^2 = 8
2x + 1 = ± √8
2x = - 1 ± √8
x = ( - 1 ± √8 ) / 2
method 2:
-8(2x+1)(2x+1)=-64
(2x+1)^2=8
4 x^2 + 4x + 1 = 8
4 x^2 + 4x - 7 = 0
x = { - 4 ± √[4^2 - 4*4*(-7)] } / (2*4)
x = ( - 4 ± √128 ) / 8
x = ( - 4 ± 8 √2 ) / 8
x = ( - 1 ± 2 √2 ) / 2
x = ( - 1 ± √8 ) / 2
2007-09-14 6:25 am
先將-8除過去,變做
(2x+1)(2x+1)=8
ps.(2x+1)(2x+1)即係(2x+1)²
so,可以將佢兩邊開方
2x+1=√8
2x= ±√8-1
x= -1±√8 /2
(2x+1)(2x+1)=8
ps.(2x+1)(2x+1)即係(2x+1)²
so,可以將佢兩邊開方
2x+1=√8
2x= ±√8-1
x= -1±√8 /2
參考: 自己嘗試
2007-09-14 6:20 am
-8(2x+1)(2x+1)=-64
(2x+1)(2x+1)=8
(2x+1)^2=8
2x+1=√8 or 2x+1= -√8
x= (-1+√8)/2 ~~ or x= (-1-√8)/2 ~~
(2x+1)(2x+1)=8
(2x+1)^2=8
2x+1=√8 or 2x+1= -√8
x= (-1+√8)/2 ~~ or x= (-1-√8)/2 ~~
收錄日期: 2021-04-13 18:52:18
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