f4 maths questions

1a) delta = 0
[-2(a-1)]^2 -4[-(b+2)^2] = 0
4(a-1)^2+4(b+2)^2=0
(a-1)^2+(b+2)^2=0

because the solution of x^2+y^2=0 must be x=y=0 for any numbers x and y ,
a-1=0 , a=1
b+2 = 0 , b=-2

(1)^999+(-2)^3 = 1-8 = -7

b)
(x-a)(x-b)=0
(x-1)(x+2)=0
x^2+x-2=0

2a)
f(x) = -(x-1)(x-2)(x-k)
f(3) = -(3-1)(3-2)(3-k) + 8 = 0
-6+2k+8=0
k=-1

b)
f(-3) : -(-3-1)(-3-2)(-3+1) = 40




2010-08-16 19:05:45 補充：
for question 2a,
f(3)= 8
-(3-1)(3-2)(3-k)=8
-6+2k=8
k=7

2b)
f(-3) :
-(-3-1)(-3-2)(-3-7) = 200

To 001, how about Question 1?

2)
a)
The equation :

(x - 1)(x - 2)(x - k) = 0

(x^2 - 3x + 2)(x - k) = 0

x^3 - 3x^2 + 2x - kx^2 + 3kx - 2k = 0

x^3 - (k + 3)x^2 + (3k + 2)x - 2k = 0

-x^3 + (k + 3)x^2 - (3k + 2)x + 2k = 0

f(x) = -x^3 + (k + 3)x^2 - (3k + 2)x + 2k

f(3) = -27 + 9(k + 3) - 3(3k + 2) + 2k

= -27 + 9k - 9k + 27 - 6 + 2k

= 2k - 6 = 8

So, k = 7

b)
The equation becomes:

f(x) = -x^3 + 10x^2 - 23x + 14 = 0

f(-3) = 27 + 90 + 69 + 14 = 200

The reminder is 200.





2010-08-16 18:34:51 補充：
2)
a)
The equation :
(x - 1)(x - 2)(x - k) = 0
(x^2 - 3x + 2)(x - k) = 0
x^3 - 3x^2 + 2x - kx^2 + 3kx - 2k = 0
x^3 - (k + 3)x^2 + (3k + 2)x - 2k = 0
-x^3 + (k + 3)x^2 - (3k + 2)x + 2k = 0
f(x) = -x^3 + (k + 3)x^2 - (3k + 2)x + 2k

2010-08-16 18:34:55 補充：
f(3) = -27 + 9(k + 3) - 3(3k + 2) + 2k
= -27 + 9k - 9k + 27 - 6 + 2k
= 2k - 6 = 8
So, k = 7
b)
The equation becomes:
f(x) = -x^3 + 10x^2 - 23x + 14 = 0
f(-3) = 27 + 90 + 69 + 14 = 200
The reminder is 200.

2010-08-16 19:00:55 補充：
I don't know it!!!!!!

You know??

You many answer!!!!!!!!!

2010-08-16 19:01:03 補充：
You may answer

2010-08-16 19:01:26 補充：
Am I Q2 correct??

2010-08-16 19:03:33 補充：
我又錯??????? @@

2010-08-17 09:19:23 補充：
1. If the roots of the equation x^2 -2(a-1)x-(b+2)^2 =0 are equal , find
(a) a^999 + b^3
(b) a quadratic equation with roots a and b
a)
roots are equal, delta = 0
4(a - 1)^2 + 4(b + 2)^2 = 0
(a - 1)^2 + (b + 2)^2 = 0

2010-08-17 09:19:27 補充：
As, (a - 1)^2 and (b + 2)^2 are not negative, they are equal to 0.
(a - 1)^2 = 0 and (b + 2)^2 = 0
a = 1 and b = -2
Then,
a^999 + b^3 = 1^999 + (-2)^3 = 1 - 8 = -7
So,
roots are a and b,
then, (x - a)(x - b) = 0
(x - 1)(x + 2) = 0
x^2 + x - 2 = 0