Couldn't you simply plug into the Quadratic Formula?
a) the diagram shows the graphs of y=x-1 and y=kx^2 where k is a positive constant.
The graphs intersect at 2 dinstinct points A and B.
Write down the quadratic equation satisfied by the x-coordinates of A and B, and hence show that k<1/4
b) Describe briefly the relashioship between the graphs of y=x-1 and y=kx^2 in each of the cases
c) Show, by using a graphical argument of otherwise that when k is a negative constant, the equation x-1=kx^2 has 2 real roots, one of which lies between 0 and 1.
I need help in the last part of the Q.
I know that to show that the equation has 2 real roots we need to show that its determinant is greater than zero ,but how do we know that one of those points are between 0 and 1??
Well your use of the Quadratic formula is incorrect. If then . Solving for x gives
Now if that means , which means .
From this we can state that due to . So that means it must be that's between 0 and 1.
It should be obvious to start with that this fraction must be positive. If not, convince yourself of it.
Now let's assume that this was . Then we would have
This is an obvious contradiction, which means that is NOT . Therefore it must be between 0 and 1.
When dealing with quadratic equations, we analyze the discriminant to find whether we have two distinct real roots, one real repeated root, or complex conjugate roots.
If the discriminant is positive, then we have two distinct real roots. The discriminant is
and with we know:
The quadratic formula tells us the roots will be:
Let's examine the root:
Multiply through by
Since this must be true, then the root we began with must be on the required interval.