F.4 math---inequalities2

1)
Since y=(k-4)(x^2) - kx + 4x + k - 1>=0, then the quadratic equation may have or not have roots.
Then, discriminant<=0 (The equation will not have 2 distinct roots since the function is non-negative)
(4-k)^2-4(k-4)(k-1)<=0
(k-4)(k-4-4(k-1))<=0
(k-4)(-3k)<=0
k(k-4)>=0
k<=0 or k>=4

2)
f(x)>=g(x)
f(x)-g(x)>=0
This means the graph of f(x)-g(x) is always greater or equal to 0, that is, the equation may have or not have roots.
discriminant<=0 (Just like Q1)
(4k)^2-4(6-5k)<=0
16k^2-24+20k<=0
-2<=k<=3/4


2009-12-26 14:09:33 補充：
係喎...Thanks nelson...

2009-12-26 14:12:04 補充：
Addition for Q1, since the graph is non-negative, the coefficient of x² has to be positive.
i.e. k-4>0 -> k>4

combined with the result of the previous part, the ans is k>4

1)
Since y=(k-4)(x^2) - kx + 4x + k - 1&gt;=0, then the quadratic equation may have or not have roots.
Then, discriminant&lt;=0 (The equation will not have 2 distinct roots since the function is non-negative)
(4-k)^2-4(k-4)(k-1)&lt;=0
(k-4)(k-4-4(k-1))&lt;=0
(k-4)(-3k)&lt;=0
k(k-4)&gt;=0
k&lt;=0 or k&gt;=4

2)
f(x)&gt;=g(x)
f(x)-g(x)&gt;=0
This means the graph of f(x)-g(x) is always greater or equal to 0, that is, the equation may have or not have roots.
discriminant&lt;=0 (Just like Q1)
(4k)^2-4(6-5k)&lt;=0
16k^2-24+20k&lt;=0
-2&lt;=k&lt;=3/4

(1) k has to be > 4 so that the parabola is always positive.