The Discriminant of Quadratic Equations

Date: 08/06/98 at 10:18:07
From: Quincy 
Subject: Quadratic equation

Will you explain why b^2 - 4ac = a triangle?

Date: 08/06/98 at 11:41:58
From: Doctor Santu
Subject: Re: Quadratic equation

Hello, Quincy!

This is all about how to solve equations such as:

   40 x^2  +  5 x  +  63 = 0

right? People learned how to solve lots and lots of equations like 
this and then, a few hundred years ago, somebody asked what do the 40, 
the 5, and the 63 have to do with it? Can we find rules about what the 
answer is for every possible combination of the three "coefficients"?  
[In our example, 40 is the x^2-coefficient, 5 is the x-coefficient, and 
63 is the constant term.] So one day, they set out to solve the 
equation:

   a x^2  +  b x  +  c   =   0.

But the answers they got depended (obviously) on a, b and c, the 
coefficients. There were several different ways the problem could turn 
out, and, most interestingly, exactly which way it went was decided by 
this famous b^2 - 4ac. So they wanted to call that formula by a name, 
and the name that was decided was "discriminant," because it 
discriminated between the various solutions.

But, as you know, in math we like to call things by single letters, 
like a, b, f, x and so on. So we use the Greek letter Delta as its 
symbol. Delta looks exactly like a triangle, but it is a Greek letter, 
not a geometric shape (just as the letter o looks like a circle to 
people who don't know what it is).

The expression ax^2 + bx + c is an interesting one. With some careful 
math, it can be split into two parts:

   a * [varying part that's never negative]  -  [fixed part]

The fixed part is Delta/[4a], so you see the connection!

If Delta is negative, the whole combination has absolutely no chance of 
being zero, so the equation ax^2 + bx + c = 0 cannot be solved. To see 
this, put the whole thing over a common denominator, and you get:

   4a^2 * [varying part, never negative]  -  Delta
   -----------------------------------------------
                        4a

If Delta is negative, -Delta will be positive, and 4a^2 is certainly 
positive, so the numerator is going to be at least as big as -Delta, 
so it can never be zero, regardless of the denominator.

If Delta is positive, there will be two different values for x at 
which the whole expression becomes zero. This is the so-called generic 
case. To see this, I have to tell you that the "varying part" is just
(x + b/(2a))^2, which takes any desired positive value exactly twice.

If Delta is zero, the only value of x that will make the expression 
zero is x = -b/(2a), so you have one solution.

A much nicer way of understanding the solution is to think of it in 
terms of graphs. You must know what the graph of the expression:

   ax^2 + bx + c

looks like, in which case the three cases become obvious. Look in your 
textbook. There should be diagrams showing you the situation. There 
will be a u-shaped curve, called a parabola. There will be a 
horizontal line that's the x-axis. Each point on the x-axis stands 
for a possible x-value. The corresponding value of ax^2 + bx + c is 
just how high the point on the parabola is from the x-axis. When the 
curve actually crosses the x-axis, the height is zero, of course, so 
the value is zero. The question then becomes does this graph cross the 
x-axis? At how many points?

The a, b, and c determine the position of the parabola. The "a" 
determines whether the parabola is arch-shaped (a is negative) or 
u-shaped (a is positive). Delta determines whether the parabola crosses 
the x-axis (Delta is positive), just touches the x-axis (Delta is zero) 
or misses the x-axis completely (Delta is negative).

Thanks for writing, and write back if you want more information

- Doctor Santu, The Math Forum
Check out our web site! http://mathforum.org/dr.math/