A visual representation of Imaginary solutions
For today’s #myfavfriday I am presenting an idea that has been percolating in my head for a while. If you want to know what a #myfavfriday is, then see Druinok’s blog here.
Learners have a devil of a time with quadratics. Afterall, there can be 2 solutions, 1 solution, or no solutions in Algebra 1, and then in Algebra 2, we come at them with the fact that those equations with no solution really do have 2 solutions after all, they are just “imaginary” (could there be a worse name for them, really? Thanks a lot Descarte.”)
But I came across a picture on some site one day, and it has stuck with me. I never bookmarked it, or wrote down the site, so it is lost to me (and I have searched hard for it) but the work blew my mind, and as I have shown it to learners, they have at least gotten a sense that the “imaginary” really does have meaning.
Let’s begin with 2 equations and graphs that are simple, straight-forward and make sense. [all images are clickable to see full size]
The equations are y = x^2 – 4x + 3 and y = x^2 – 4x + 4.
A simple change of one number changes the number of solutions from 2 distinct to 2 repeating solutions, and learners don’t have a problem with that idea, generally. Then comes this bad boy.
Now they have to do the whole Quadratic formula on it to get the solution, and the solution has those i thingies in it, which makes them all confused and irritated until they wrap their heads around it. And why does it still have 2 solutions? It doesn’t touch at all!
But wait! We can play a game with this quadratic function. What if we reflect the parabola around the vertex in the downward direction? Then we end up with something that looks like this:
To do this reflection, we first had to complete the square on the original equation to get y = (x-2)^2 + 2. Now, with this equation, we can put the – sign into the equation and get the reflection, y = -(x-2)^2 + 2.
But hold on, see those 2 points where it crosses the X-axis? And see the Axis of Symmetry that goes through both equations? If we use those three points as definitions for a circle, we get the following graph and equation.
Guess what the solution to the quadratic equation y = x^2 –4x + 6 is. If you guessed 2 + root(2)i and 2 – root(2)i then you are absolutely correct.
The real number part of the complex solution of a quadratic with two imaginary roots is the X value of the Axis of Symmetry, and the imaginary part of the solution is the radius of the circle created by the center and endpoints created when the inverted parabola crosses the X-Axis!
Okay, mind blown. Why? How could I prove this?
Aha! now come into play the hours I spend on a motorcycle every summer. How could I PROVE that this will always work? I have the proof. I am working it up, but it is a pain to type. That, I think, will be the focus of a future, #myfavfriday!
[And I really need to look in to a LaTex module for my blog if I am going to do math. The equations look horrible.]
Edit: 29 July 2012: I proved this assertion, at least to my satisfaction in a followup post: http://blog.mrwaddell.net/archives/348
Edit: 4 August 2012: I found, stuffed in the bottom of my backpack, a rumpled piece of paper with this link on it. I think I did this page justice with my treatment. I wish I had found the page before I spent hours thinking about how to prove it, it gives the suggestion right there at the bottom!
From: Teaching Mathematics, 2nd edition by m. Sobel and E. Maletsky