一元二次

水龍蜥

2x^2+2p=(4p+1)x
求p,使公式有equal real roots
教下我ok?

sapphire

hv u learn the eqt. D=b^2-4ac yet? in fact this question is extremely easy after u learn this.

2x^2+2p=(4p+1)x
2x^2-(4p+1)x+2p=0  which is a quad. eqt. in x

the eqt. has equal real root if and only if D=0
(4p+1)^2-4(2)(2p)=0
solve the quad. eqt. , u will get the ans p=1/4

水龍蜥

原帖由 sapphire 於 11/9/2007 07:24 PM 發表
hv u learn the eqt. D=b^2-4ac yet? in fact this question is extremely easy after u learn this.

2x^2+2p=(4p+1)x
2x^2-(4p+1)x+2p=0  which is a quad. eqt. in x

the eqt. has equal real root if and  ...

忘記了用交叉法..難怪計不到

ccshong28

原帖由 水龍蜥 於 11/9/2007 08:48 PM 發表

忘記了用交叉法..難怪計不到

我係F.2既，想問下咩係交叉法= ='
我係咁諗既：
2x^2+2p=(4p+1)x
2x^2=4px+x-2p
2x^2-x=4px-2p
x(2x-1)=2p(2x-1)
x=2p

When x=2p,
2x^2+2p=(4p+1)x
2(2p)^2+2p=(4p+1)x
4p^2+4p^2+2p=(4p+1)x
4p^2+4p^2+2p=(2p+1)(2p)
4p^2+4p^2+2p=4p^2+2p
4p^2=2p
2p=1
p=1/2

我有冇計錯= ='

[ 本帖最後由 ccshong28 於 11/9/2007 09:39 PM 編輯 ]

K龍

原帖由 ccshong28 於 11/9/2007 09:38 PM 發表

我係F.2既，想問下咩係交叉法= ='
我係咁諗既：
2x^2+2p=(4p+1)x
2x^2=4px+x-2p
2x^2-x=4px-2p
x(2x-1)=2p(2x-1)
x=2p

When x=2p,
2x^2+2p=(4p+1)x
2(2p)^2+2p=(4p+1)x
4p^2+4p^2+2p=(4p+1)x
4p^2+4 ...

when x = 2p
2x^2 + 2p = (4p + 1)x
2(2p)^2 + 2p = (4p + 1)(2p)
2(4p^2) + 2p = 8p^2 + 2p
8p^2 + 2p = 8p^2 + 2p
結果是什麼也沒得出

要活用"have equal real root"這info，一般來就都是都以求D的方法去計

sapphire

apart from the reply above

2x^2-x=4px-2p
x(2x-1)=2p(2x-1)
x=2p

here u should not divide both sides by (2x-1)
always remember that u should not divide both sides by a number with an unknown as the number divided may equal to zero

[ 本帖最後由 sapphire 於 11/9/2007 10:31 PM 編輯 ]

ccshong28

原帖由 sapphire 於 11/9/2007 10:24 PM 發表
apart from the reply above

here u should not divide both sides by (2x-1)
always remember that u should not divide both sides by a number with an unknown as the number divided may equal to zero

翻參考書，明白了
thx~^^:D