Exam Finished~

抹茶可樂

Ｄ
=4(a-b)^2-4(c-a)(b-c)
=4(a-b)^2+4(a-c)(b-c)
=4(a^2-2ab+b^2+ab-ac-bc+c^2)
=4(a^2+b^2+c^2-ab-ac-bc)
=2(2a^2+2b^2+2c^2-2ab-2ac-2bc) <+整個問題的重點
=2(a^2-2ab+b^2 + a^2-2ac+c^2 + b^2-2bc+c^2)
=2[(a-b)^2+(b-c)^2+(c-a)^2]
>0 (if a=b=c, Ｄ=0)

細薯

Originally posted by 落雷 at 2005-3-6 09:02 PM:
Ｄ
=4(a-b)^2-4(c-a)(b-c)
=4(a-b)^2+4(a-c)(b-c)
=4(a^2-2ab+b^2+ab-ac-bc+c^2)
=4(a^2+b^2+c^2-ab-ac-bc)
=2(2a^2+2b^2+2c^2-2ab-2ac-2bc) <+整個問題的重點
=2(a^2-2ab+b^2 + a^2-2ac+c^2 + b^2-2bc ...

thx ar~~
我就係去到

=4(a^2+b^2+c^2-ab-ac-bc)

ng識去重點果part

=2(2a^2+2b^2+2c^2-2ab-2ac-2bc) <+整個問題的重點

thx~
[E-Hap]

ps: Thursday A.Maths. Exam.....!

細薯

help~
========================================
New Trend A.Maths. (P.29)
9.
Suppose
   _
  |   y=x+m
-|                                       where m is a constant.
  |_ y=x^2 + 4x - 3m

(a)If P(x1,y1) and Q(x2,y2) are their points of intersection,form a quadratic equation with roots x1 and x2.

(b)If P and Q coincide,find the value of m.
========================================

[ Last edited by smallpotato226 on 2005-3-9 at 05:45 PM ]

抹茶可樂

Originally posted by smallpotato226 at 2005-3-9 02:51 PM:
help~
========================================
New Trend A.Maths. (P.29)
9.
Suppose
   _
  |   y=x+m
-|                                       where m is a constant.
  |_ y=x^2 + 4x - 3m
...

As x1 and x2 are solutions,

eliminating y ,
x+m=x^2+4x-3m
x^2+3x-4m=0


Ｄ=0
9-4(-4m)=0
9+16m=0
m=-9/16

細薯

Originally posted by 落雷 at 2005-3-9 04:57 PM:

As x1 and x2 are solutions,

eliminating y ,
x+m=x^2+4x-3m
x^2+3x-4m=0


Ｄ=0
9-4(-4m)=0
9+16m=0
m=-9/16

i don't understand why this equation has the solutions x1,x2.....
x^2+3x-4m=0