How do you determine whether there are two, one or no real solutions given the graph of a quadratics function does not have an x-intercept?

Sep 8, 2017

If a graph of a quadratic, $f \left(x\right)$, does not have an x-intercept
then $f \left(x\right) = 0$ has no Real solutions.

Explanation:

The x-axis is composed of all points for which $f \left(x\right)$ (or, if you prefer, $y$) is equal to $0$

If the graph of $f \left(x\right)$ does not have an x-intercept
then it has no (Real) points for which $f \left(x\right) = 0$

Sep 8, 2017

See explanation

Explanation:

No x-intercept means that it does not cross the x-axis. Thus two solutions is definitely ruled out.

However, if you DO NOT INCLUDE a point of coincidence (Vertex coincides with the x-axis) in the phrase "does not have an x-intercept". Then there could be a single value solution if you equate the quadratic to 0. Some people say that it still has two in such a case but they are both the same value. I do not like this way of thinking!

On the other hand, if you DO INCLUDE a point of coincidence in the phrase, then the plot does not cross the x-axis nor does any point on the curve coincide with it. In such an interpretation there is NO SOLUTION THAT IS REAL.