f.4 maths,求救!!!!!!!!!!!!!!!!!!

1. (a+bi)(3-5i)
=3a-5ai+3bi+5b
=(3a+5b)+(3b-5a)i
3a+5b=-9
3b-5a=-19
by solving, a=2,b=-3

2.(a)
the length=16-x
A=x(16-x)

(b)
x(16-x)=55
-x^2+16x-55=0
x=5 or x=11

(c) x(16-x)
=-x^2+16x
=-(x-8)^2+64
so that the maximum value of A is 64

3.(a+1)+(b+1)=-3 a+b=-5
(a+1)(b+1)=-2 ab+a+b+1=-2 ab-5+1=-2 ab=2

a^2+b^2=(a+b)^2-2ab =25-4=21
(a^2)(b^2)=(ab)^2=4

4.係唔係f(x)=(x-a)(x-b)(x+1)-3?
f(1)=1
1=(1-a)(1-b)(1+1)-3
4=(1-a)(1-b)(2)
(1-a)(1-b)=2
(a-1)(b-1)=2

b. a=3,b=2 or a=2,b=3 because 2=2x1

2014-03-13 19:47:39 補充：
(a and b is +ve integer) so that can not be (-1)x(-2)

1. If (a + bi)(3 - 5i) = -9 -19i, find the value of a and b.

　(a + bi)(3 - 5i)
= 3a - 5ai + 3bi + 5b
= (3a + 5b) + (3b - 5a)i

Comparing real parts and imaginary parts.
{ 3a + 5b = -9
{ 3b - 5a = -19

{ 15a + 25b = -45
{ 9b - 15a = -57

Adding the two equations gives
34b = -102
b = -3

3a - 15 = -9

2014-03-13 19:33:52 補充：
3a = 6
a = 2

{ a = 2
{ b = -3


2.There is a rectangle with perimeter 32 cm and width x cm. Let A cm² be its area.

(a) Express A in terms of x.

Length = 32/2 - x = (16 - x) cm

Area = A cm² = x cm × (16 - x) cm

A = x(16 - x) = -x² + 16x

(b) If the area is 55 cm², find the possible value(s) of x

2014-03-13 19:34:11 補充：
A = 55
-x² + 16x = 55
x² - 16x + 55 = 0
(x - 5)(x - 11) = 0
x = 5 or x = 11

(c) What is the maximum value of A.

A = -x² + 16x
　= -(x² - 16x)
　= -(x² - 16x + 64 - 64)
　= -(x² - 16x + 64) + 64
　= -(x - 8)² + 64

The maximum value of A is 64.

(The corresponding value of x is 8.)

2014-03-13 19:34:28 補充：
3. Given that α+1 and β+1 are the roots of the quadratic equation
　　 x² + 3x - 2 = 0

(a) Find the values of β² + α² and (α²)(β²).

It is known that
{ (α+1) + (β+1) = -3
{ (α+1) × (β+1) = -2

{ α + β = -5
{ αβ + α + β + 1 = -2

{ α + β = -5
{ αβ - 5 + 1 = -2

{ α + β = -5
{ αβ = 2

2014-03-13 19:34:39 補充：
Consider β² + α²
= (β + α)² - 2αβ
= (-5)² - 2(2)
= 25 - 4
= 21

(α²)(β²) = (αβ)² = 2² = 4

2014-03-13 19:35:25 補充：
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