a math question

1.a)
z(x - 3) = (x - 2)^2
x^2 - 4x + 4 = zx - 3z
x^2 - (z + 4)x + 3z + 4 =0
Since x is real,
△ = (z + 4)^2 - 4(3z + 4) >= 0
z^2 - 4z >= 0
z(z - 4) >= 0
The range of possible values of z:
z <= 0 or z >= 4
b)
Range of x such that y is defined:
x - 2 > 0 and x - 3 > 0
x > 3
y = 2log(x - 2) - log(x - 3) = log[(x - 2)^2 / (x - 3)]
10^y = (x - 2)^2 / (x - 3)
10^y (x - 3) = (x - 2)^2
By (a):
10^y <= 0 (rejected) or 10^y >= 4
y >= log4
Minimun value of y = log4
2.a)
Since α is the common root of the two equations,
3α^2 + aα + b = 0 ... (1)
3α^2 + bα + a = 0 ... (2)
(1) - (2): (a - b)α= a - b
Since a and b are distnct rational numbers,
α= 1
b)
Since α= 1, from (1) in (a):
3 + a + b = 0
a + b = -3
Since a and b are the roots of x^2 + hx + k = 0,
h = -(a + b) = 3
k = ab = -a(a + 3) = -(a^2 + 3a)
As a is a rational number, let a = p/q, where p and q are intergers and (p, q) = 1.
k = -(p^2/q^2 + 3p/q) = -p(p + 3q)/q^2
Since k is a positive interger,
p(p + 3q) is divisible by q^2 and hence by q,
As (p, q) = 1, p + 3q is divisible by q,
thus it implies p is divisible by q,
we must have q = 1, that is a is an interger.
a + b = -3
k = ab >0
By trials and errors, (a = -2 and b= -1) or (a = -1 and b= -2)
k = (-2)(-1) = 2

3α^2+aα+b=0_____________(1)
3α^2+bα+a=0_____________(2)

將α=1代入(1),(2)
再解可得a=-2 and b=-1

2009-06-28 22:33:51 補充：
第2題:
http://f.imagehost.org/0519/ScreenHunter_02_Jun_28_16_31.gif

1.a)let z(x-3)=(x-2)^2 . if x is real, find the range of possibl values of z.

solution:
z(x-3)=(x-2)^2
zx-3z=x^2-4x+4
x^2+(z-4)x+(4-3z)=0
since x is real , so b^2-4ac>0
(z-4)^2-4(4-3z)>0
z^2-8z+16-16+12z>0
z^2+4z>0
z>0 or z<-4//

b)let y=log(x-2)-log(x-3).find the range of x such that y is defined. under the above range, find the minimun value of y

the defined range x>3

y=log(x-2)-log(x-3)
=[ln(x-2)-ln(x-3)]log e
dy/dx= log e [1/(x-2)-1/(x-3)]
dy/dx=0
=>1/(x-2)-1/(x-3)=0
=>x-3-x+2=0
(in fact , in this case , no solution)

2.let α be the common root of the two equations
3x^2+ax+b=0 , 3x^2+bx+a=0
where a and b are distnct rational numbers.

a) since the α is the root of both equations ,
so 3α^2+aα+b=o and 3α^2+bα+a=0
(1)-(2): (a-b)α+(b-a)=0
=>α=1//

b) a=-2 and b=-1 ,

then , h= -(a+b)=-(-2-1)=3 and k=ab=(-1)(-2)=2//






2009-06-28 17:26:32 補充：
that is one and only one matching .

2009-06-28 17:29:43 補充：
y=2log(x-2)-log(x-3)

y=log (x-2)^2/ (x-3)

since the number in log must be greater than 0

so that (x-2)^2/ (x-3) > 0 and x-3 <>0

=> x>2 and x<>3

use the same method ,

dy/dx= log e [ 2/(x-2) - 1/(x-3)]

dy/dx=0

2009-06-28 17:32:55 補充：
2/(x-2) - 1/(x-3)=0

=>2x-6-x+2=0

=>x=4//

and dy^2/dx^2= log e [1/(x-3)^2-2/(x-2)^2] >0

so the minimum value of that is y= log 4//