reversing coefficients in a quadratic

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apple2357

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I noticed something when i solved the following two quadratics:

x^2-5x+6 =0, gives x=2, 3

and

6x^2-5x+1 =0 gives x=1/2, 1/3

( It came out of a problem to do with alpha, beta, sum of roots, product of roots problem!)

The second quadratic is made by reversing the coeff in the first one and the solutions appear to be the reciprocals! This is initially surprising but i was able to prove this by looking at two quadratics (x-k)(x-m) and (x-1/k)(x-1/m).

Three main
questions:

1. Is there a geometric way of seeing why this is works? Is there some kind of transformation taking place?
2. Playing on graphical software and this appears to work for all quadratics ( except when x=0 is a solution), not just factorised ones? Is this provable?
3. Does this work for higher order polynomials?

Any thoughts, as this is completely novel to me!
 
Given the equation \(\displaystyle ax^2+ bx+ c= 0\), divide by \(\displaystyle x^2\) to get \(\displaystyle a+ \frac{b}{x}+ \frac{c}{x^2} = c\left(\frac{1}{x}\right)^2+ b\left(\frac{1}{x}\right)+ a= 0\). Yes, you can do the same thing with any polynomial.
 
HallsofIvy said:
Given the equation \(\displaystyle ax^2+ bx+ c= 0\), divide by \(\displaystyle x^2\) to get \(\displaystyle a+ \frac{b}{x}+ \frac{c}{x^2} = c\left(\frac{1}{x}\right)^2+ b\left(\frac{1}{x}\right)+ a= 0\). Yes, you can do the same thing with any polynomial.
Click to expand...
Remember to stay away from analyzing situations where x = 0
 
You need to be careful when you have an equation of the form ax^2 + bx. When you switch the equation around you get 0x^2 +bx + a= bx+a Not bx^2 + ax. Also as Khan has already pointed out, you need to be careful when x=0 is a root. When x=0 is a root what happens with the reversed quadratic? There are many interested things going on here. If you go into this deeply you can give a low level talk on this, perhaps at the Mathematical Association of Two-Year Colleges in your state (assuming you live in the US)

What you observed just begs the question does this happen with higher (and lower) polynomials? Will using Halls' method ( now you are famous!) can you arrive at the same conclusions you got with the quadratics or possibly by another method. Maybe it only works with even degree polynomials?
See where you get with this. Please show us your work.
 
Subhotosh Khan said:
Remember to stay away from analyzing situations where x = 0
Click to expand...
NO!!, let the OP analyze what happens when x =0 and how it affects the reversed quadratic!!!! Come on, this is a teaching site! Now go to the corner for 1/0 minutes.
 
Jomo said:
NO!!, let the OP analyze what happens when x =0 and how it affects the reversed quadratic!!!! Come on, this is a teaching site! Now go to the corner for 1/0 minutes.
Click to expand...

To be fair I did recognise some problem with x=0 , I wrote it in my original post. I am interested in a graphical interpretation and that’s what I am going to look at next.
 
I created a geogebra file which allows me to play with the problem:

1578427867484.png

And here is the case where one of the roots is zero!

1578427929601.png

Are we saying there is an infinite hidden solution somewhere or does the problem just become meaningless?
 
What happened here is that the "reversed" quadratic is not quadratic, but linear. That much is clear. And it is to be expected, because one of the zeros does not exist. (But don't call it a "solution"; we're talking about a function, which has zeros, not an equation, which has solutions.)

If you want, you can think of it as if the linear function were a quadratic in which one zero has "moved off to infinity". That's what you see as you make c closer and closer to zero, right? Of course it can't really be that, but it's a degenerate quadratic.

So yes, in a sense one zero is "hidden" way off to the left or right (it doesn't matter which). Just don't say that to a mathematician ...
 
Thanks. Good correction on the misuse of solutions rather than zeros!
I guess i was thinking about solutions to the quadratic=0 but i really meant the zeros.
 
Does this result hold for complex numbers too? Can't see why it wouldn't but feels weird.
 
Last edited:
apple2357 said:
Thanks. Good correction on the misuse of solutions rather than zeros!
I guess i was thinking about solutions to the quadratic=0 but i really meant the zeros.
Click to expand...
Actually, looking back at your original question, I was wrong; you were talking about equations. I was focusing only on the graphs you'd just drawn.
apple2357 said:
Does this result hold for complex numbers too? Can't see why it wouldn't but feels weird.
Click to expand...
Find out for yourself! Pick an equation with complex solutions (I assume you aren't talking about complex coefficients, but you could consider that, too, if you don't mind a lot of work!), and see what happens. Graphing won't help, of course, which would be part of the reason for the weirdness.

Then try applying post #2 to this case, and see if you can still prove it.
 
I realized if you used the quadratic formula for both equations

(-b + sqrt(b^2 - 4 a c))/(2 a)×(-b - sqrt(b^2 - 4 a c))/(2 c)


This conveniently simplifies to 1.
So i think it will work for complex solutions too.
 
HallsofIvy said:
Given the equation \(\displaystyle ax^2+ bx+ c= 0\), divide by \(\displaystyle x^2\) to get \(\displaystyle a+ \frac{b}{x}+ \frac{c}{x^2} = c\left(\frac{1}{x}\right)^2+ b\left(\frac{1}{x}\right)+ a= 0\). Yes, you can do the same thing with any polynomial.
Click to expand...
Is there anything there that requires that a, b, c, or x be real numbers?
 
apple2357 said:
I realized if you used the quadratic formula for both equations

(-b + sqrt(b^2 - 4 a c))/(2 a)×(-b - sqrt(b^2 - 4 a c))/(2 c)


This conveniently simplifies to 1.
So i think it will work for complex solutions too.
Click to expand...
Now that is interesting. Does anyone see that this is obvious? Can you please explain why?
 
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