williamkhw

IF 4x^2-(1+m)x+1=0 has two realroots,find the value of m

引用:

原帖由 williamkhw 於 2007-10-13 07:43 發表
IF 4x^2-(1+m)x+1=0 has two realroots,find the value of m

since the equation has 2 real roots
so D=0
(-(1+m))^2-4(4)(1)=0
1+2m+4m^2-8=0
4m^2+2m-7=0
m=(-2+-(2^2-4x4(-7))^1/2)/2x4
m=(-1+-(1)(29)^1/2)/4

仲有...你果題真係就咁寫住2 real root?
係2equal real root吧?
還是前面有分題- -
因為如果係2real root 個D應該大過0or=0
but你要搵value so me當你係2equal real root 黎做- -

[ 本帖最後由 『小龍龍﹏ 於 2007-10-13 08:02 編輯 ]

williamkhw

m=(-2+-(2^2-4x4(-7))^1/2)/2x4
m=(-1+-(1)(29)^1/2)/4

呢度唔明...呢個係咩來的= =?

姐果個2次方程ge公式呢- -
(x)^1/2咪姐=x ge開方--
我5識整開方個圖

williamkhw

引用:

原帖由 『小龍龍﹏ 於 2007-10-13 13:23 發表
姐果個2次方程ge公式呢- -
(x)^1/2咪姐=x ge開方--
我5識整開方個圖

-b+-開方b^2-4ac   /2a?

嗯- -
係lie條式la
姐識做啦=]