F4 A.Maths discriminant

Fisrt, (k-2)x^2 + (2k-1)x + (k-5) <-3
Then both sides +3 , we get (k-2)x^2 + (2k-1)x + (k-2) <0
The value of x should alsways be less then -3. i.e. after +3, less than 0. That means the graph of y = f(x) should not touch the x-axis.
Therefore, discriminent < 0.
Plus, k < 2 because the value is less than 0 as mentioned above. That means the graph of y = f(x) opens downwards. So coefficient of x < 0, i.e. k - 2 < 0, k < 2

2007-10-13 20:32:27 補充：
(b) find also the range of values of k for which f(x) is less than -3.This is what the question requires.And according to your answer, value of x(roots of the equation?) less than 0 means the graph should not touch the x-axis. Why can&#39;t?

2007-10-13 20:33:57 補充：
Because if the graph has any intersection point(s) with the x-axis, at that intersection point we have x = 0, not x 0.Maybe having an example.Consider the graph y = -(x - 1)(x - 2).We know that the graphs open downwards and the roots are 1 and 2.

2007-10-13 20:35:32 補充：
So, for 1

2007-10-13 20:36:54 補充：
is less than x and x is less than 2, we have f(x) &gt; 0But we need to have f(x) &gt; 0 for all x.Therefore, the graph should not touch the x-axis. In other words, discriminent &lt; 0