InfoJunkie4Life
Jun 22, 2010, 10:56 AM
Maybe if we all put our minds together we can have a mathematical breakthrough.
Here's my dilemma. Back in high school, I had just switched schools so I couldn't be in any of the available classes for a whole semester. I got my hands on an intermediate algebra book and started reading through it. I figured this would be a good precursor to my precalculas class coming up.
I had passed the NYS Regent Math B exam with decent scores, but there were still so many details left out of my training. I came across polynomials in my reading and thought "This Looks Easy." The first thing I came across was:
\frac{-b \pm \sqrt(b^2-4ac)}{2a}
I had seen it before, as I am sure all of you have. Then I found the proof for it in the next few days, I will try to reproduce it here.
ax^2 + bx + c = 0
subtract c
ax^2 + bx = -c
divide by a
x^2 + \frac{b}{a}x = -\frac{c}{a}
complete the squre
x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}
Simplify
(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}
Square Root of both
x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}
X by itself:
x = \frac{-b \pm \sqrt(b^2-4ac)}{2a}
Now given all of this, why can't we do the same thing for 3rd order polynomials.
ax^3 + bx^2 + cx + d = 0
ax^3 + bx^2 + cx = -d
x^3 + \frac{b}{a}x^2 + \frac{c}{a}x = -\frac{d}{a}
At this point I get lost. I would think you would take an x out of the left, however, that doesn't do much. You would think that there would be something you can add to both sides without an x in it so that it can be factored. I've done a bit of reading on this and they say there are proofs proving that it is impossible to solve this. The only thing I can think of is goofing around with imaginary numbers and what not. I can't think of anything to add to both sides of the equation to complete the "square."
x(x^2 + \frac{b}{a}x + \frac{c}{a}) = -\frac{d}{a}
Here's my dilemma. Back in high school, I had just switched schools so I couldn't be in any of the available classes for a whole semester. I got my hands on an intermediate algebra book and started reading through it. I figured this would be a good precursor to my precalculas class coming up.
I had passed the NYS Regent Math B exam with decent scores, but there were still so many details left out of my training. I came across polynomials in my reading and thought "This Looks Easy." The first thing I came across was:
\frac{-b \pm \sqrt(b^2-4ac)}{2a}
I had seen it before, as I am sure all of you have. Then I found the proof for it in the next few days, I will try to reproduce it here.
ax^2 + bx + c = 0
subtract c
ax^2 + bx = -c
divide by a
x^2 + \frac{b}{a}x = -\frac{c}{a}
complete the squre
x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}
Simplify
(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}
Square Root of both
x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}
X by itself:
x = \frac{-b \pm \sqrt(b^2-4ac)}{2a}
Now given all of this, why can't we do the same thing for 3rd order polynomials.
ax^3 + bx^2 + cx + d = 0
ax^3 + bx^2 + cx = -d
x^3 + \frac{b}{a}x^2 + \frac{c}{a}x = -\frac{d}{a}
At this point I get lost. I would think you would take an x out of the left, however, that doesn't do much. You would think that there would be something you can add to both sides without an x in it so that it can be factored. I've done a bit of reading on this and they say there are proofs proving that it is impossible to solve this. The only thing I can think of is goofing around with imaginary numbers and what not. I can't think of anything to add to both sides of the equation to complete the "square."
x(x^2 + \frac{b}{a}x + \frac{c}{a}) = -\frac{d}{a}