A.Maths. Exam

細薯

help!
tomolo A.Math Exam la~~
i have some question don't know how to do!~

help~~~~~~~~~~~

1.Convert the expresion x^2-4n+1 into the form (x+a)^2 +b and write down the values of a and b in terms of n.

[ Last edited by smallpotato226 on 2004-10-27 at 04:12 PM ]

Ronald0077

落雷  在 2004-10-29 10:46 PM 發表:

Let f(x) = (16-k)x^2 +12x -k , where k is a constant.
a) Find the discriminant of f(x)=0 .
b) Find the condition of k such that f(x)≧0 for all real values of x .

a) Δ = b² - 4ac
       = 12² - 4(16-k)(-k)
       = 144 + 64k - 4k²

b) As f(x) ≧ 0, Δ ≦ 0 and 16-k ＞ 0
                      144 + 64k - 4k²≦0 and k＜16
                      k² - 16k - 36 ≧ 0 and k＜16
                      (k - 18)(k + 2) ≧ 0 and k＜16
                      (k ≦ -2 or k ≧ 18) and k＜16
                   ∴k ≦ -2

see if sth wrong @@~

細薯

fish  在 2004-10-29 10:49 PM 發表:

afriad＜－－－－afraid ....."

改了.......(成日錯><)

fish

smallpotato226  在 2004-10-28 01:06 PM 發表:

don't be afriad~
i have typed the correct question!

does anyone have the A.Maths Textbook of the name called "New Trend Additional Mathematics Volume One"??
my question is come from  ...

afriad＜－－－－afraid ....."

抹茶可樂

Let f(x) = (16-k)x^2 +12x -k , where k is a constant.
a) Find the discriminant of f(x)=0 .
b) Find the condition of k such that f(x)≧0 for all real values of x .

細薯

落雷  在 2004-10-27 07:01 PM 發表:

I'm afraid you of wrong typing of queation

don't be afraid~
i have typed the correct question!

does anyone have the A.Maths Textbook of the name called "New Trend Additional Mathematics Volume One"??
my question is come from P.6 , Classwork 5

++++++++++++++++++++++++++++++++++++++++++++++
testing

5<SUP>2</SUP>

haha~~funny~qq~XDD

[ Last edited by smallpotato226 on 2004-10-30 at 09:27 AM ]

Echo21

itphone

抹茶可樂

I'm afraid you of wrong typing of queation
x^2 -4nx +1
=x^2 -2(x)(2n) +1
=x^2 -2(x)(2n) +(2n)^2 -(2n)^2+1
=(x-2n)^2+1-4n^2

a=-2n
b=1-4n^2

Echo21

剛剛看了最上樓的題目：
1.Convert the expresion x^2-4n+1 into the form (x+a)^2 +b and write down the values of a and b in terms of n.

答案（未必對）：
     x^2 - 4nx + 1
= (x^2 - 4nx) + 1
= [x^2 - 2(2nx)] + 1
= [x^2 - 2(2nx) + 4n^2] - 4n^2 + 1
= (x - 2n)^2 - 4n^2 + 1

Sub. (x - 2n)^2 - 4n^2 + 1 into (x + a)^2 + b
∴ a = -2n//
　 b = -4n + 1//

完成！

[ Last edited by Echo21 on 2004-10-27 at 07:30 PM ]