Let's discuss Mathematics

@Samson , about the +- presentation, another thought of how it can be presented using not just the real number line but distance (absolute value). So Xo is used as a pivot point to max/min x, which turns it (in this world) into an analogue of 0. For the special case, Xo is 0. I included also one use within secondary education math. Other uses include limits (by just having an inequality instead of an equality).

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A question. The following is my proof that if you have two equations of the form:
ax^2+bx+c
cx^2+bx+a
with the coefficients having the same value (c,a,b) and just c and a changing positions,
also with c(a) not zero

then it follows (a) that of their two roots -as c(a) is not zero, there are two roots and none is zero-, one root of the first will be the reciprocal of one root of the second
and also (b) that actually both of their roots will be reciprocals of the other roots.

I wish to ask:
1) is this true? (that always both roots of the first will have a reciprocal in roots of the second)
2) can you think of a different way to prove this? (eg with identities to do with b and the root of the discriminant)
3) I suspect that the reciprocal is always one with altered position in regards to -b/coefficient of the stable term being added or subtracted from the square root of the discriminant - eg if x1=(-b+sqrootD)/2a, it will always be the reciprocal of x2'=(-b-sqrootD)/2c. Is this true?

Here are my solutions:

1738769720238.png


Thanks for any help :) Eg @a pen-dragon
 
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This is indeed true and there is an easier (and more general) proof, providing a similar statement for any polynomial:

We want to solve a*x²+b*x+c=0. (Or the roots of an arbitrary polynomial)
Let r be the reciprocal of x. Then x=1/r and we obtain:
a*(1/r)²+b*(1/r)+c=0
If we multiply this by r² we get our final result:
c*r²+b*r+a=0

The condition that a and c must not be equal to zero emerges from demanding r to be finite and non-zero. a or c equal to zero will always produce a root at zero.

More generally this procedure will always reverse the order of the coefficients.


Concerning 3), this is indeed true, as can easily be checked by multiplying the standard solution of the roots. Note the order of the signs in the first terms is exchanged. Both cases result in the same.
(-b+-\sqrt(b²-4ac))/(2a)*(-b-+\sqrt(b²-4ac))/(2c) = (b²-(b²-4ac))/4ac = 1
 
This is indeed true and there is an easier (and more general) proof, providing a similar statement for any polynomial:

We want to solve a*x²+b*x+c=0. (Or the roots of an arbitrary polynomial)
Let r be the reciprocal of x. Then x=1/r and we obtain:
a*(1/r)²+b*(1/r)+c=0
If we multiply this by r² we get our final result:
c*r²+b*r+a=0

The condition that a and c must not be equal to zero emerges from demanding r to be finite and non-zero. a or c equal to zero will always produce a root at zero.

More generally this procedure will always reverse the order of the coefficients.


Concerning 3), this is indeed true, as can easily be checked by multiplying the standard solution of the roots. Note the order of the signs in the first terms is exchanged. Both cases result in the same.
(-b+-\sqrt(b²-4ac))/(2a)*(-b-+\sqrt(b²-4ac))/(2c) = (b²-(b²-4ac))/4ac = 1
Thanks. For some bizarre reason I didn't bother with the direct multiplication to find a proof for question 3 :D
And yes, the proof for question1 is certainly more condensed and alludes to more general use too - although I constructed my own meaning to use it for question3...
 
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