f.4 inequality, absolute value

(3)
a)x>=3 & x>=-3 => x>=3
b)3＞x＞0 & x>1 => 3>x>1
c)0＞x, the result calculated should be 1>x (the sign is incorrect)
0>x & 1>x => 0>x
All three cases together => x>1 or x<0

2009-07-05 21:56:09 補充：
Q1: http://img90.imageshack.us/img90/2528/67017089.jpg
Q2: http://img183.imageshack.us/img183/7570/92374392.jpg
Q3: http://img90.imageshack.us/img90/3356/60167509.jpg
Q4: http://img90.imageshack.us/img90/38/46195213.jpg

Yahoo!知識+管理員 , 你說final ans of 3. is x<0 or x>1 , 這是3c的ans嗎?

If m = -4, the equation becomes x^2 - (-4 + 3)x + [ 2 - (-4)] = 0
that is x^2 + x + 6 = 0, which has no real roots. But the question said alpha and beta are real roots?? So m = - 4 is really the answer?

4.
x^2+(k-2)x-(k-1)=0
(x-1)[x-(1-k)]=0
x=1 or x=1-k
Since︱α︳=︱β︳
︱1-k︱=1
k=0 or k=2

2009-07-04 16:20:07 補充：
︱α︳=︱β+1︳
α=β+1 orα=-(β+1)
α-β=1 or α+β=-1
(α-β)2=1 orα+β=-1
(α+β) 2-4αβ=1 orα+β=-1
(m+3)2-4(2-m)=1 or m+3=-1
m2+6m+9-8m+4=1 or m=-4
m2-2m+12=0 or m=-4
no real root or m=-4


2. a) x>0
x>(3/x)+2
x2>3+2x
x2-2x-3>0
(x+1)(x-3)>0
∴x<-1 or x>3
b) x<0
-x>-(3/x)+2
-x2>-3+2x
x2+2x-3<0
(x-1)(x+3)<0
∴-3<x<1

3. a)x＞=3
(x-3)/2x<1
(x-3)<2x
-x<3
∴ x>-3
b)3＞x＞0
-(x-3)/2x<1
-(x-3)<2x
-x+3<2x
3<3x
∴ x>1
c)0＞x
-(x-3)/2x<1
-(x-3)<2x
-x+3<2x
3<3x
∴x>1

4.

α2=β2
(α+β)(α-β)=0
(1) α+β=0
-(k-2)=0
k=2
(2) α-β=0
(α-β)2=0
(α+β) 2-4αβ=0
(2-k)2-4(k-1)=0
k=0

2009-07-04 19:19:05 補充：
更正:
(m+3)^2-4(2-m)=1 or m+3=-1
m^2+6m+9-8+4m=1 or m=-4(rejected.)
m^2+10m=0
m(m+10)=0
m=0 or m=-10

2009-07-04 19:19:23 補充：
更正:
(m+3)^2-4(2-m)=1 or m+3=-1
m^2+6m+9-8+4m=1 or m=-4(rejected.)
m^2+10m=0
m(m+10)=0
m=0 or m=-10

2009-07-04 19:56:42 補充：
2. a)
x>(3/x)+2 and x>0
x^2>3+2x and x>0
x^2-2x-3>0 and x>0
(x+1)(x-3)>0 and x>0
(x<-1 or x>3) and x>0
∴x>3

b) x<0
x>(3/x)+2 and x<0
x^2<3+2x and x<0
x^2-2x-3<0 and x<0
(x+1)(x-3)<0 and x<0
-1<3 and x<0
∴-1<0

1.if α, β are the real roots of the equation x^2-(m+3)x+(2-m)= 0 and if ︱α︳=︱β+1︳, find the value of m
α+β=m+3__________(1)
αβ=2-m_____________(2)
From (1),β=m+3-α____(3)
From ︱α︳=︱β+1︳,
︱α︳=︱m+3-α+1︳
︱α︳=︱m-α+4︳
α=m-α+4 or α= -(m-α+4)
2α=m+4 or α= -m+α-4
m=4-2α or m= -α+α-4
thus, m=4-2α or m= -4

2. solve x>(3/x)+2 for each of the following cases:
a) x>0
x>(3/x)+2
x2>3+2x
x2-2x-3>0
(x+1)(x-3)>0
thus, x<-1 or x>3
b) x<0
-x>-(3/x)+2
-x2>-3+2x
0>x2-3+2x
x2+2x-3<0
(x-1)(x+3)<0
thus, -3<x<1

3. solve the inequality(︱x-3︳/2x)<1 by considering each of the following cases:
a)x＞3
(x-3)/2x<1
(x-3)<2x
-x<3
thus, x>-3
b)3＞x＞0
-(x-3)/2x<1
-(x-3)<2x
-x+3<2x
3<3x
thus, x>1
c)0＞x
-(x-3)/2x<1
-(x-3)<2x
-x+3<2x
3<3x
thus, x>1

4. α, β are the real roots of the equation x^2+(k-2)x-(k-1)=0
if ︱α︳=︱β︳, find k
α+β=k-2 _____________(1)
αβ=-(k-1)___________(2)
From (1), β=k-2-α____(3)
From︱α︳=︱β︳,
︱α︳=︱k-2-α︳
α=k-2-α or α= -(k-2-α)
2α+2=k or α= -k+2+α
k=2α+2 or k= -α+2+α
thus, k=2α+2 or k= 2