Inequalities

Solution

For kx^2+4x+k>3
kx^2+4x+k-3>0
kx^2+4x+ (k-3) >0
Hence coefficient of x^2>0
∴k>0

Also, Δ< 0
4^2 – 4 (k) (k -3) < 0
16 – 4k (k – 3) < 0
4 –k (k – 3) < 0
4 –k^2 + 3k < 0
0 < k^2 – 3k -4
k^2 – 3k -4 > 0
(k + 1) (k – 4) > 0
∴ k<-1 or k > 4

Since k>0,
k<-1 must be rejected

∴ Overall solution: k > 4

2006-10-17 01:18:21 補充：
&lt means less then> means greater than

When kx^2 + 4x + k > 3 for all real values of x,
kx^2 + 4x + k - 3 > 0
Discriminant < 0, k > 0
4^2 - 4(k)(k-3) < 0
16 - 4k^2 + 12k < 0
(-4k^2 + 12k +16) / -4 > 0 / -4
k^2 -3k -4 > 0
(k - 4)(k + 1) > 0
k < -1 (rej.) or k > 4 because k > 0
Therefore the answer is k > 4.

P.S. I think the system cannot present out the signs greater than and smaller than. This is what it has shown out: > = greater than, < = smaller than

kx^2+4x+k>3
kx^2+4x+k-3>0
DISCRIMINANT<0
16-4k(k-3)<0
16-4k^2+12k<0
4k^2-12k-16>0
k^2-3k-4>0
(k-4)(k+1)>0
k<-1 or k>4