# algebra

• Dec 20th 2008, 06:05 AM
chogo
algebra
i have this question

its just for fun, i DONT want the answer just want someone to tell me what approach is good for such questions

so

a+b+c=6

0<a<b<c

ab+bc+ca=9

prove 0<a<1<b<3<c<4

------

i just started on it and i got these ideas

ab should be < b

and ab< both ac and bc
• Dec 20th 2008, 06:28 AM
running-gag
Hi

One possible approach is to consider a, b and c as solutions of a 3rd degree polynomial equation

We know that if a, b and c are solutions of
$x^3 + \alpha x^2 + \beta x + \gamma = 0$
then
$\alpha = - (a + b + c)$
$\beta = ab + bc + ca$
$\gamma = - abc$

a, b and c are therefore solutions of
$x^3 -6 x^2 + 9x + \gamma = 0$ with $\gamma < 0$

Study the variations of the function $f(x) = x^3 -6 x^2 + 9x + \gamma$ and especially the conditions to have 3 roots and you will get your answer
• Dec 20th 2008, 06:43 AM
chogo
see this is like an 'interview question' which someone told me about

and it should be solved in a minute with pen and paper

so im trying to find a simple answer

thanks tho
• Dec 20th 2008, 06:51 AM
chogo
also forgive my ignorance, but if you do assume a cubic polynomial, how do you know the solutions for alpha,beta and gamma are those you showed. how did you get those
• Dec 20th 2008, 07:02 AM
running-gag
Because if a, b and c are solutions of
$x^3 + \alpha x^2 + \beta x + \gamma = 0$
then
$x^3 + \alpha x^2 + \beta x + \gamma = (x-a) (x-b) (x-c)$

Develop the second member and you will find
$\alpha = - (a + b + c)$
$\beta = ab + bc + ca$
$\gamma = - abc$
• Dec 20th 2008, 07:04 AM
running-gag
Quote:

Originally Posted by chogo
see this is like an 'interview question' which someone told me about

and it should be solved in a minute with pen and paper

so im trying to find a simple answer

thanks tho

My method allows to find the solution in a minute with pen and paper (Wink) (Nod)
• Dec 20th 2008, 08:03 AM
chogo
ok i did some reading, and i now understand the number of solutions cant exceed the degree

alpha beta and gamma are the solutions and they are such because of the degrees on the member

so you get the equation you gave me. Now how should i solve that, i could put it through matlab, but that cheating :)
• Dec 20th 2008, 08:52 AM
running-gag
Maybe my advice is not relevant because I do not know which is your level in mathematics

a, b and c being solutions of
$x^3 -6 x^2 + 9x + \gamma = 0$ with $\gamma < 0$

if you study the variations of the function $f(x) = x^3 -6 x^2 + 9x + \gamma$ by finding the derivative, you will see that f :
is increasing from 0 to 1 with f(0) = -abc < 0 and f(1) = 4 - abc which must be > 0
is decreasing between 1 and 3 with f(1) = 4 - abc > 0 and f(3) = -abc < 0
in increasing between 3 and 4 with f(3) = -abc < 0 and f(4) = 4-abc > 0

Therefore one of the solutions is between 0 and 1, one is between 1 and 3 and one is between 3 and 4
• Dec 20th 2008, 09:11 AM
franckherve1
curious
and what values do u consider for gamma? isn't there a way to find it?
• Dec 20th 2008, 12:08 PM
running-gag
Here is the graphs of the function $f(x) = x^3 -6 x^2 + 9x + \gamma$ for several values of $\gamma$ between -5 and 1

The value of $\gamma$ has no impact on the form of the curve because no impact on the derivative, it is only a translation on y axis

You can see that $\gamma$ must be between -4 and 0 in order to have 3 roots and in this case the roots are between 0 and 1, between 1 and 3 and between 3 and 4

http://imageshack-france.com/out.php/i273001_Graph2.JPG
• Dec 20th 2008, 05:06 PM
chogo
thanks for the help

one last question

any recommended topics i should read to further understand this?
• Dec 20th 2008, 05:38 PM
chogo
and when i open (x-a)(x-b)(x-c)

i get the coeffs as

alpha=-c
beta=ab+bc
and gamma = -abc

is that now how i get the members?

sorry for taking up so much of your time. dont bother to reply if you dont have time, ill try and figure it out for myself
• Dec 21st 2008, 02:24 AM
running-gag
Don' t worry !
(x-a)(x-b)(x-c) = (x²-ax-bx+ab)(x-c)
(x-a)(x-b)(x-c) = (x²-(a+b)x+ab)(x-c)
(x-a)(x-b)(x-c) = x^3-cx²-(a+b)x²+(a+b)c+abx-abc
(x-a)(x-b)(x-c) = x^3-(a+b+c)x²+(ac+bc+ab)x-abc