Quadratic Equations (Motivational) : Discriminants and Imaginary (Complex) Numbers - with examples

QUESTION
Quadratic Equations : Discriminants and Imaginary (Complex) Numbers
When using the quadratic formula to solve a quadratic equation $ax^2 + bx + c = 0$, the discriminant is $b^2 - 4ac$. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, $\sqrt{-1} = i$. ) 

Explain what the value of the discriminant means to the graph of $y = ax^2 + bx + c$. Hint: Chose values of $a$, $b$ and $c$ to create a particular discriminant.  Then, graph the corresponding equation. 

What are examples of quadratic equations or applications with imaginary numbers?

SOLUTION =====================
So we consider the quadratic function $y=ax^2+bx+c$. 

When $a>0$, the parabola is concave up and has a minimum turning point when $x=-{b\over 2a}$. 
When $a<0$, the parabola is concave down and has a maximum turning point at  $x=-{b\over 2a}$. 

The intersection points (if any) with the x-axis occur when $y=0$, so when $ax^2+bx+c=0$. 

When $a>0$ so the parabola curves upwards. 
In this case, if the discriminant is positive,  this means that the intersection of the parabola with the $x$-axis gives two solutions, so it crosses the $x$-axis twice. 
If the discriminant is zero,  this means that the intersection of the parabola with the $x$-axis gives precisely one solution, so it just touches the $x$-axis. 
If the discriminant is negative, the equation $ax^2+bx+c=0$ has no real solutions (only imaginary ones) so this means that the parabola does not cross the $x$=axis at all. (it is what they call positive definite) 

So to summarise the situation. Let's denote the discriminant by $\Delta = b^2-4ac$ 

(1)(i)  $a>0, \Delta>0$ - the parabola curves up, and crosses the x-axis in two places 
(1)(ii)  $a>0, \Delta=0$ - the parabola curves up, and touches the x-axis in exactly one place 
(1)(iii)  $a>0, \Delta<0$ - the parabola curves up, and never crosses the x-axis (positive definite) 

The case for $a<0$ is similar, 

(2)(i)  $a<0, \Delta>0$ - the parabola curves down, and crosses the x-axis in two places 
(2)(ii)  $a<0, \Delta=0$ - the parabola curves down, and touches the x-axis in exactly one place 
(2)(iii)  $a<0, \Delta<0$ - the parabola curves down, and never crosses the x-axis (negative definite) 

Please see graph for examples. 

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EXAMPLE 1: A Stones Throw
When you throw a stone up into the air it makes an arc, and guess what the shape of that arc is? The stone is an example of a projectile. 
It's a downward parabola with some equation say $y= ax^2+bx+c$ where $a<0$. 

Now say you were asked to calculate whether your stone would reach a certain height $H$. 

In a nutshell, you solve this equation for $x$,  $ax^2+bx+c=H$  or  $ax^2 + bx + (c-H) = 0$.  This is another quadratic equation.  If the discriminant is negative this means that the stone will not reach height $H$ at all!  So getting imaginary solutions does give you information about a physical problem!

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EXAMPLE (1):(Isaac Newton looked at this one) 

One problem could be the motion of the 
pendulum . This motion can be described in terms of a differential equation, and in 
the case of small swings of the pendulum this equation can be solved to find the time of the swing. Solving it 
requires finding the solution to a quadratic equation! 
If x is the angle of swing of the pendulum, then Newton realised that there were numbers a,b and c which 
depend on such features as the length of the pendulum, air resistance and the strength of the gravitational 
force so that the differential equation describing the motion was 
  $ax''+bx'+cx=0$.  where $x' = dx/dt$ and $x'' = d^2x/dt^2$ 
           
Here t is time, $x''$ is the acceleration of the pendulum and  $x'$ is its velocity. 

The mathematician Leonhard Euler devised a means of solving this particular equation that relied on 
the solution of a quadratic equation. Euler suggested the existence of a solution of the form 
$x(t) = exp(wt) $

Substituting into the differential equation and dividing by $e^(wt)$  gives the following equation for $w$: 
$aw^2 + bw + c = 0$. 
This is very familiar! All we need to do to solve the original differential equation is to solve this quadratic 
equation and substitute back for w. By doing this we can accurately predict the behaviour of the pendulum. 


NOW, getting more to the point of imaginary solutions applications yoiu asked for in your problem. 
If discriminant =$b^2 - 4ac > 0$ , then the quadratic equation has two real solutions and physically this one corresponds to a pendulum with a lot of friction (or a pendulum moving in a liquid such as water) 
. 
In contrast, if discriminant $=b^2 - 4ac < 0$ , then the same differential equation has oscillating solutions which look  more like the motions of the pendulum that we are familiar with. 

REFERENCE:  this document has more background and also some graphs of the above solutions. 
http://plus.maths.org/issue30/features/quadratic/feat.pdf 
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This method of developing a differential equation like a $y''+by'+c=0$ and getting an 'auxiliary' quadratic equation $aw^2+bw+c=0$ occurs often in many other  physics/engineering applications.