Amaths - Harder Problems of Inequalities

(1) Find the range of real values of k if x^2 + 2(1+k) + (3+k) >_ (larger or equal to) 0 for all real values of x.

The equation has no solution or one repeated root [not delta>=0 as when the equation >0,it means that the graph is always above the x-axis and no interception between the graph and x-axis]
delta<=0
4(1+k)^2-4(3+k)<=0
4(1+2k+k^2)-12-4k<=0
4k^2+4k-8<=0
k^2+k-2<=0
(k+2)(k-1)<=0
-2<=k<=1
2. Find the range of real values of k if -2x^2 + 3x - (3k+1) < 0 for all real values of x.
[the graph never intercept with the x-axis]
delta<0
3^2-4(-2)[-(3k+1)]<0
9-8(3k+1)<0
9-24k-8<0
1<24k
k>1/24
3.Find the range of real values of k if (3k-1)x^2 - 2x + (k-1) <_ 0 for all real values of x.
delta<=0 [sinilar as (1)]
4-4(3k-1)(k-1)<=0
4-4(3k^2-4k+1)<=0
-12k^2+16k<=0
12k^2-16k>=0
k(3k-4)>=0
k<=0 or k>=4/3
combine with (3k-1)<0 ==>k<1/3
(k<=0 or k>=4/3) and k<1/3
k<=0
[以前個2題唔使係因為開口方向同k冇關,唔使理,但呢題有關,因為(3k-1)必須<0先會開口向下,相反(3k-1)大過)會導致開口向上,圖像會與x軸有交點]


2007-12-10 16:47:39 補充：
如上，不等號使5使調轉??? e.g.-a -b當乘或除以一個負數去另一邊時,不等號就要轉!!

2007-12-11 13:32:35 補充：
如果圖像y=f(x)大於或小於0,即係同x軸冇交點,delta =0

For the equations to have roots, b2 - 4ac &gt;= 0 *Quadratic Equation*

1. b=2(1+k) *I guess you mistype it and should be 2(1+k)x
a = 1
c = (3+k)

[2(1+k)]^2 - 4 (1) (3+k) &gt;= 0
4 [ k2+2k+1] - 12 - 4k &gt;= 0
4k2 + 8k +4 -12 -4k &gt;= 0
4k2 + 4k -8 &gt;= 0
k2 + k - 2 &gt;= 0
(k+2)(k-1) = 0
k = -2 and +1
If k=0, (k+2)(k-1) &lt; 0
Therefore range of k: k&lt;= -2 and k &gt;= +1 for all real values of X

2. b= 3
a= -2
c= -(3k+1)

3^2 - 4 (-2) [-(3k+1)] &gt;=0
9 - 8(3k+1) &gt;= 0
9 - 24k - 8 &gt;= 0
1 &gt;= 24k
1/24 &gt;= k for all real values of x

3. b= -2
a= (3k-1)
c= (k-1)

(-2)^2 - 4 (3k-1) (k-1) &gt;= 0
4 - 4 (3k-1) (k-1) &gt;= 0
1 - (3k-1) (k-1) &gt;=0
1 - [ 3k2 -4k + 1] &gt;= 0
-3k2 - 4k &gt;= 0
3k2 + 4k &lt;= 0
k(3k + 4) &lt;= 0
k=0 , k= -4/3
If k=1 or k=-1, k(3k+4) not less than 0

Therefore, -4/3 &lt;= k &lt;= 0 for all real values of X