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標題: Quadratic Equation [打印本頁]

作者: ting 時間: 23/9/2006 01:56 PM
標題: Quadratic Equation
If a, b and c are real numbers and not all equal, prove that the quadratic equation
(c-a)x^2-2(a-b)x+(b-c)=0 has unequal real roots.

計不到
開始對自己適不適合讀理科產生懷疑

作者: kazuhiko 時間: 23/9/2006 05:40 PM
discriminant
= [2(a-b)]^2 - 4(c-a)(b-c)
= 4(a-b)^2 - 4(c-a)(b-c)
= 4[ a^2 - 2ab + b^2 - (bc - c^2 - ab + ac)]
= 4( a^2 + b^2 + c^2 - ab - bc - ac )
= 2[(a-b)^2 + (b-c)^2 + (c-a)^2] > 0 (a, b and c are real numbers and not all equal)

Therefore, the quadratic equation (c-a)x^2-2(a-b)x+(b-c)=0 has unequal real roots

[ 本帖最後由 kazuhiko 於 24/9/2006 21:39 編輯 ]

作者: Royal 時間: 24/9/2006 10:45 AM
那是附加數抑或是普通數學?
我沒有修附加數, 但有學類似的東東,也計不到xd

作者: ting 時間: 24/9/2006 11:47 AM
amths來的
計到4( a^2 + b^2 + c^2 - ab - bc - ac )就卡住了
想不到開學第一堂amath(沒硬性把amath和maths分開)已有問題
灰到爆開

[ 本帖最後由 (ABC)ting 於 24/9/2006 12:01 編輯 ]

作者: 魔術師Chikorita 時間: 24/9/2006 12:03 PM

原帖由 kazuhiko 於 23/9/2006 05:40 PM 發表
determinant
= ^2 - 4(c-a)(b-c)
= 4(a-b)^2 - 4(c-a)(b-c)
= 4
= 4( a^2 + b^2 + c^2 - ab - bc - ac )
= 2 > 0 (a, b and c are real numbers and not all equal)

Therefore, the quadratic equat ...

a=1, b=2, c=3的時候已經錯了......

作者: ting 時間: 24/9/2006 12:17 PM

原帖由 魔術師Chikorita 於 24/9/2006 12:03 發表

a=1, b=2, c=3的時候已經錯了......

wt do u "up"?

who tell u a=1 , b=2, c=3.
You assume that?
a, b, c are only for discriminant.
They are constant terms.
When u substitute a=1 b=2 c=3, x is still a variable.
How could u know it is wrong?
I think u can only assume 2 of them in order to find the range of the last one

[ 本帖最後由 (ABC)ting 於 24/9/2006 12:29 編輯 ]

作者: 魔術師Chikorita 時間: 24/9/2006 12:32 PM

原帖由 (ABC)ting 於 24/9/2006 12:17 PM 發表

wt do u "up"?
who tell u a=1 , b=2, c=3.
You assume that?
a, b, c are only for discriminant.
They are constant terms.
When u substitute a=1 b=2 c=3, x is still a variable.
How ...

又學多一個東東了

作者: 流映 時間: 24/9/2006 12:38 PM
唔需要灰心…呢條唔識是正常的

作者: Royal 時間: 24/9/2006 12:59 PM

原帖由 kazuhiko 於 23/9/2006 17:40 發表
determinant
= ^2 - 4(c-a)(b-c)
= 4(a-b)^2 - 4(c-a)(b-c)
= 4
= 4( a^2 + b^2 + c^2 - ab - bc - ac )
= 2 > 0 (a, b and c are real numbers and not all equal)

Therefore, the quadratic equat ...

係唔係(a-b)^2一定會係正數架?

**估唔到amaths真係咁難...好彩冇修

作者: ting 時間: 24/9/2006 02:27 PM

原帖由 Royal 於 24/9/2006 12:59 發表

係唔係(a-b)^2一定會係正數架?

**估唔到amaths真係咁難...好彩冇修

yes
square must be positive

作者: sapphire 時間: 24/9/2006 05:05 PM

原帖由 (ABC)ting 於 24/9/2006 14:27 發表

yes
square must be positive

but not for complex number
u will study this in F.6 Pure Math

作者: ting 時間: 24/9/2006 05:41 PM

原帖由 sapphire 於 24/9/2006 17:05 發表

but not for complex number
u will study this in F.6 Pure Math

難怪老師在說的時候加了一句「就form4而言」XD

作者: JOHNLAM17 時間: 24/9/2006 06:57 PM

原帖由 (ABC)ting 於 24/9/2006 14:27 發表

yes
square must be positive

erm..., I think there is a small mistake

In general, for a term ---> (a-b)² (Assuming that a and b are real numbers)
It can either be a positive number orzero.(when a = b)

But in this question, however,
because a, b, c are not equal,
so the discriminant must be positive

[ 本帖最後由 JOHNLAM17 於 24/9/2006 18:59 編輯 ]

作者: ting 時間: 24/9/2006 07:49 PM

原帖由 JOHNLAM17 於 24/9/2006 18:57 發表

erm..., I think there is a small mistake

In general, for a term ---> (a-b)² (Assuming that a and b are real numbers)
It can either be a positive number or zero.(when a = b)

But in ...

o yeah
u a right

作者: Royal 時間: 25/9/2006 12:59 PM
今天做maths有類似的問題，類似的方法
但當然冇咁複雜啦

我老師話ce maths連續十幾年都冇出過類似題目
(因為難度直迫amaths)
但係難保一日hkeaa會發癲出呢d罕見題目
例如今年ce出過absolute error(係十幾年o黎第一次出)
<<雖然呢個topic好似係f.2既...但係都有好多人錯

另外, square唔一定係positive
但一定唔係negative number(即係可以係0, 可以係positive)

[ 本帖最後由 Royal 於 25/9/2006 13:00 編輯 ]

作者: 流映 時間: 25/9/2006 08:28 PM

原帖由 Royal 於 25/9/2006 12:59 發表
例如今年ce出過absolute error(係十幾年o黎第一次出)

因為以前個syllabus沒有
但今年是新的syllabus
所以咪有出

作者: ting 時間: 3/10/2006 08:33 PM
Let Q be the common root of the two equations
3x^2+ax+b=0
and 3X^2+bx+a=0
where a and b are distinct rational numbers
(a) find the value of Q (solved, Q=1)
(b)If a and b are the roots of x^2+hx+k=0, where h and k are positive integers, find the value of h and k (I have found h, h=3)

Do you know how to find k?
Thx

作者: kazuhiko 時間: 4/10/2006 12:11 AM
a)

Q is the common root of the two equations

Therefore,
3Q^2+aQ+b=0 ......(1)
and 3Q^2+bQ+a=0 ......(2)

(1) - (2): (a-b)Q + (b-a) = 0
Q= 1

b)

put Q=1 into (1), we have
a+b+3 = 0
b= -(a+3)

a and b are the roots of x^2+hx+k=0
Therefore,
h = -(a+b) and k = ab

h = -(a+b) = -[a - (a+3)] = 3
k = ab = -a(a+3) > 0
=> -3 < a < 0
Since k is a positive integer, a must be equal to -1 or -2
which both gives k=2

Remarks
Actually, (a,b) = (-1,-2) or (-2,-1) gives the same two equations in part (a)

[ 本帖最後由 kazuhiko 於 4/10/2006 00:17 編輯 ]

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