Quadratic Equation

sapphire

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

ting

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

JOHNLAM17

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 ]

ting

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

Royal

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 ]

流映

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

ting

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

kazuhiko

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 ]