Exam Finished~

細薯

Later i will post the questions and discuss with you all~~
please help me also~~
thx a lot~~^^

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New Trend A.Maths. Volume One
Chapter 1 ,P.14 ,Ex.1B,Q.16

16.(a)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.

    (b)If a = 1, b = 3,find the discriminant of the equation in (a).Hence find the range of values of c so that the equation has unequal real roots.
============================================

[ Last edited by smallpotato226 on 2005-3-10 at 05:50 PM ]

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

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

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 ]

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.....!

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

one A.Maths. question in #1,plx help~~

細薯

用D=b^2-4ac 都ng得.........><
========================================
呀!我諗到D野丫!
掉轉泥calculate 都得wor~~

TRY:
(a)
since x must be integral solutions,
therefore,sub. x = 1,2,3 into the equation satisfy the equation

x^2 + ax + 12 =0

1+a+12=0    =>   a=-13

4+2a+12=0  =>  a=-8

9+3a+12=0   => a=-7

etc.

(Give up to three answers if necessary.)

ok ma??

1)G.Maths.
In each of the following equations,what integral value(s) should a take so that the equation will have integral solutions?
(Give up to three answers if necessary.)
(a)x^2 + ax + 12 =0
(b)x^2 + 9x + a =0
(i have answer,plx teach me the steps~~thx~~)

[ Last edited by smallpotato226 on 2005-2-16 at 10:13 AM ]

kokoro

a同b係2題來的…
諗係好易 XD
Present就煩 XD

suny

係唔係二元一次方程?
係唔係咁.....


┌x^2 + ax + 12 =0   (i)
└x^2 + 9x + a =0     (ii)

從(i),
x^2+ax+12=0
      x^2+ax= -12
        x(x+a)= -12
            x+a= -12/x
                a= (-12/x)-x   (iii)

(iii)代入(ii),可得
x^2+9x+(-12/x)-x=0
     x^2+8x-(12/x)=0
      x(x+8) -(12/x)=0
                 x(x+8)=12/x
                     x+8=12/x^2
                         x=(12/x^2)-8


跟住........唔係好掂[E-Swe]