a little discovery while fooling around with 20X^2 - 60X= 0

allegansveritatem

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I was fooling around with a problem the other day and came up with this formulation: 20X^2 - 60X= 0. I know this can be solved easily by factoring out a 10 and seeing immediately what X had to be in the result: 10 (2X^2 - 6X) = 0. But for some reason I didn't do this. Instead I just started inserting numbers I knew the squares of. Eventually I can to 3 and noticed two thing. 1: It satisfied the equation and 2: It could be obtained by dividing the coefficient of the first degree variable with the coefficient of the same variable squared. I tried this with other constants and found that no matter what I used as coefficients, I could always get the value of X by dividing the first degree variable's coefficient with the coefficient of the same variable to the second degree. This surprised me because I haven't come across this short cut (if I can call it that) anywhere in my big book. Is anyone familiar with this phenomenon or have I hit on something original that I can copyright and make a fortune on?

(Please excuse the typos in this post. My word processing is screwed up temporarily and I don't remember which key to press to correct the problem)
 
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I was fooling around with a problem the other day and came up with this formulation: 20X^2 - 60X= 0. I know this can be solved easily by factoring out a 10 and seeing immediately what X had to be in the result: 10 (2X^2 - 6X) = 0. But for some reason I didn't do this. Instead I just started inserting numbers I knew the squares of. Eventually I can to 3 and noticed two thing. 1: It satisfied the equation and 2: It could be obtained by dividing the coefficient of the first degree variable with the coefficient of the same variable squared. I tried this with other constants and found that no matter what I used as coefficients, I could always get the value of X by dividing the first degree variable's coefficient with the coefficient of the same variable to the second degree. This surprised me because I haven't come across this short cut (if I can call it that) anywhere in my big book. Is anyone familiar with this phenomenon or have I hit on something original that I can copyright and make a fortune on?

(Please excuse the typos in this post. My word processing is screwed up temporarily and I don't remember which key to press to correct the problem)

Nice, but I don't think you will make a million bucks out of it.

Consider this:

\(\displaystyle ax^2 - bx =0\)

\(\displaystyle ax^2 = bx\)

Note that \(\displaystyle x=0\) is clearly a solution.

To find the other (non-zero) solution, divide both sides by \(\displaystyle x\), which you can do since it is not zero, to get:

\(\displaystyle ax =b\)

\(\displaystyle x = \frac{b}{a}\)
 
Nice, but I don't think you will make a million bucks out of it.

Consider this:

\(\displaystyle ax^2 - bx =0\)

\(\displaystyle ax^2 = bx\)

Note that \(\displaystyle x=0\) is clearly a solution.

To find the other (non-zero) solution, divide both sides by \(\displaystyle x\), which you can do since it is not zero, to get:

\(\displaystyle ax =b\)

\(\displaystyle x = \frac{b}{a}\)

yes, there is a lot of beauty in that one too. I think mine is just a bit more elegant in that you go right to the jugular without balancing the sides so much. Well, I guess it is a matter of taste.
 
yes, there is a lot of beauty in that one too. I think mine is just a bit more elegant in that you go right to the jugular without balancing the sides so much. Well, I guess it is a matter of taste.
Harry knows that. He is just proving it for you. And no, your method is not more elegant at all as you did not prove that it works. Harry proved it and now can use it for the rest of his life knowing it always works. You at best can say that it will probably work on all problems because it worked for the few that you tried. Real, elegant math is showing why it works!
By the way, you can factor out 20X, not just 10.
 
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