what does a non real solution suggest when using the quadratic equation? Explain how the graph of a quadratic equation/ function with non real solutions differs from those graphs with real solutions.

• What have you tried? Have you grabbed a plotting package and looked at some examples? – vonbrand Apr 28 '14 at 19:13

First note that a non-real solution implies there is another one as they always come in pairs.
Graphs with of quadratic with real solutions will cross the $x$-axis. If it doesn't it will not cross the $x$-axis.
In the equation $f(x) = ax^2 + bx + c$, consider the determinant $\Delta = b^2 - 4ac$. See the picture to see all the possibilities that the sign of $\Delta$ can take.

In summary :
- $\Delta > 0$ implies two real roots
- $\Delta = 0$ implies one repeated real root
- $\Delta < 0$ implies roots are not real

The line you are interested in the yellow on which doesn't cross the $x$-axis line.

• An explanation on the downvote would be greatly appreciated. – user88595 Apr 28 '14 at 19:03

If there is no real solution then the graph does not cross the x-axis.

A non-real solution suggests that there is no real value of $x$ which will give a point that lies on your function. A real solution means that there is some real number $c$ where $=c$ and the points at which $x$ will equal to $c$ will cross the $x$- axis.

So $\Delta$ $>$ $0$ means there are two real solutions exactly where the blue parabola crosses the $x$- axis(you can see the two bold points).

At $\Delta$ $=$ $0$, you can see that the parabola only touches the $x$- axis so there will be two real solutions here but they will be the same value.

Finally, at $\Delta$ $<$ $0$, you can see that there are no points at which the parabola intersects the $x$- axis. The parabola will go on without bound in the upward direction. So $x$ can't have any value.

Remember these graphs are assumed to lie on the real plane. Had this been a complex plane the picture would look different.So basically real solutions are $x$-values and if the function doesn't cross the $x$-axis there cannot be roots for that graph.