Jeff Geha - 50 tips question (1 Viewer)

[Premus]

hey

in tip 28 of jeff geha's book.....i dont understand how you do example 3 part ii)
im not certain how they find the equation....
can you pls explain what you need to substitute instead of 'x' to get the desired equation and how you get it?

Thanks!

 

[Harimau]

You'll probably get more help if you just post the question instead of the question number.

 

[Premus]

sorry...

the constants p and q are such that the equation x^3 + px + q = 0 has three real non-zero roots, a,b,c

find the equation whose roots are 1/ (a^2b^2) , 1/ (b^2c^2) , 1 / (a^2 c^2)

Thanks

 

[Rorix]

by inspection, sub x= sqrt(y/(q^2))

 

[Premus]

can i ask how you got that?

 

[Archman]

hmm, after a bit more inspection, i think u should sub x = sqrt(y * q^2) instead
then express the polynomial in y and thats what you want.

 

[Rorix]

after 2nd inspection, i agree.

 

[mojako]

after some more extra inspection, Rorix's corrected solution at 08:13 PM Thursday is still wrong.

after even more inspection and working backwards, Ive concluded that Archman sub sub x = sqrt(y * q^2) because:
the new roots are 1/(a^2 b^2), 1/(b^2 c^2), 1/(a^2 c^2)
which can all be represented by
c^2/(abc)^2, a^2/(abc)^2, b^2/(abc)^2
which are
c^2/q^2, a^2/q^2, b^2/q^2, using the product of roots formula (or whatever you want t call it)

remembering a, b, and c are x values, and
letting the new polynomials be in y, just to make things different, its roots will be
y = x^2/q^2
and go from there

 

[CrashOveride]

can i ask how you got that?

Basically you want to let y = x^2 / (abc)^2 . This is so for every root of the original equation (a,b,c) , we now form the roots for our new equation that fit the requirement 1/a^2.b^2 etc (sub in to see what i mean). Noting that abc is the product of the roots ( =-q) we have y=x^2/q^2 so you re-arrange for x , sub that in and make it a polynomial in y.

 

[nit]

lol @ Crash...Mojako beat you there mate

 

[mojako]

lol @ Crash...Mojako beat you there mate

???
I'm confused by what you said... beat what?
Crash was just explaining the process.
In invisible text: basically there are two individuals in me. Anything done by one individual is not related to, and is not known by, the other.

 

[Slidey]

Upon inspection, I question: how does the invisible individual know that he doesn't know the visible individual?

 

[mojako]

The invisible indivual knows the visible individual.
But, the invisible individual doesn't know what the visible individual will do, is doing, did do and has done.

 

[Rorix]

My head hurts.







I'm going to go drink beer and watch rugby.