|
![](/i/fill.gif) |
Orchid XP v8 wrote:
> I still vividly remember the day I figured out why that famous formula
> actually solves quadratic equations. Maybe I'll share it with you?
OK, so "everybody knows" (?!) that ax^2 + bx + c = 0 can be transformed
into the magical expression
x = (-b +/- Sqrt(b^2 - 4ac)) / 2a
But why is that, exactly? Well, looking in my sister's A-level maths
book [unlike me, my sister was *taught* mathematics] I discover a long
sequence of algebraic manipulations that slowly transforms the one into
the other.
Well, that proves it *works*, but *why* does it work??
One day, I figured it out. Watch this:
The (linear) polynomial (x - A) has one solution: A. And the quadratic
polynomial (x - A)(x - B) has two solutions: A and B. So here we just
*constructed* a quadratic with *known* solutions!
(This may seem completely unremarkable, but remember I have no
mathematical training. It seemed pretty impressive to me!)
Now, if we expand that out, we get
x^2 - Ax - Bx + AB = 0
and with some rearranging,
x^2 + (-A-B)x + (AB) = 0
If we match that up against
x^2 + bx + c = 0
then we have
b = -A-B
c = AB
Or, put another way, -b = A+B. In other words, if we take
ax^2 + bx + c = 0
and divide through a, we now know the sum and product of the solutions!
But how do we figure out the solutions themselves given only their sum
and product?
Well, given that -b = A+B, there are an infinite number of possible
solutions. Similarly, c = AB on its own has infinitely many solutions.
Hopefully there is only two solutions to *both* equations. But what are
they? Hmm.
The next revelation: Since -b = A+B, then -b/2 = (A+B)/2. In other
words, the arithmetic mean of A and B. So -b/2 is a point exactly half
way between the two solutions. If I can just figure out how far apart
they are, I can compute them!
Clearly you can't determine the seperation from the sum. It's not
obvious that you can determine it from the product either though. But
that product must come into the equation somehow. Hmm.
Of course, I already know what the final solution is, just not why. That
gave me a clue.
So what we have is A+B, but what we *want* is A-B. Well now, what
happens if we square these?
(A+B)^2 = A^2 + 2AB + B^2
(A-B)^2 = A^2 - 2AB - B^2 (A mathematician would know this off by
heart, but it took me a while to work out.)
Ooo, that's *interesting*... The difference between the two is only in
the product in the middle! And we *have* that product!
From here, it's just a matter of tying up loose ends. b^2 = (A+B)^2.
The difference between sum^2 and difference^2 is exactly 4AB, so we have
(A-B)^2 = (A+B)^2 - 4AB = b^2 - 4c
Since we want to add/subtract half the difference to half the sum, we have
x = -b/2 +/- Sqrt(b^2 - 4c)/2
= (-b +/- Sqrt(b^2 - 4c))/2
Ah, but we already eliminated "a", didn't we? Can we put that right back
into the main formula? Well, comparing with the "standard" formula and
shifting a little algebra, it becomes clear that we can.
x = (-b +/- Sqrt(b^2 - 4ac))/2a
QED.
Bouyed up by this, I immediately set about performing the same trick for
the cubic. Obviously, this did not end well.
However figured out that you can compute the difference from the sum and
the product... GENIUS!! o_O I would never have known in a million years...
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
Post a reply to this message
|
![](/i/fill.gif) |