Further maths. Cubics. What?

djmccabie · #1 $Old$ October 3rd, 2008, 05:28 AM

Ok this 1 confused my maths teacher! You might need a pen and paper for this....

We have the equation 3x^3 + 6x^2 - 4x +7 = 0 with roots a,b,c
we gotta find the equation with root b+c, c+a, a+b
We did quite a lot of it this is how far we got...

written in the form x^3 - (sum)x^2 + (pairs)x - (product) = 0

sum = -6/3 = -2
pairs = -4/3
product = -7/3

New sum = (b+c) + (c+a) + (a+b)
= 2a + 2b + 2c
= 2(a+b+c)
=2*-2 = -4 We know this from the SUM worked out earlier

New Pairs = (b+c)(c+a) + (c+a)(a+b) + (b+c)(a+b)
= bc + ab + y^2 + ac + ac +bc + a^2 + ba + b^2 + ba + bc + ac
= a^2 + b^2 + c^2 + 3(cb + ca + ab)
= (a+b+c)^2 - 2(ab + ac + bc) + 3(-4/3) we know this from the PAIRS worked out earlier
=(-2)^2 - 2(-4/3) + (-4)
= 4 + 8/3 - 4
=8/3

Here comes the challenge...

New product = (b+c)(c+a)(a+b)
=(bc + ba + c^2 + ac)(a+b)
= bca + yb^2 + ba^2 + ab^2 + ac^2 + bc^2 + ca^2 + abc
= 2abc + a^2(b+c) + c^2(a+b) + b^2(c+a)
= 2(-7/3) .............

Please Help!!

ThePerfectHacker · #2 $Old$ October 3rd, 2008, 06:02 AM

Quote:

Originally Posted by djmccabie $View Post$

Ok this 1 confused my maths teacher! You might need a pen and paper for this....

We have the equation 3x^3 + 6x^2 - 4x +7 = 0 with roots a,b,c
we gotta find the equation with root b+c, c+a, a+b
We did quite a lot of it this is how far we got...

It would be $[x-(b+c)][x - (a+c)] [x - (a+b)] = 0$ .

Note, $(b+c)+(a+c)+(a+b) = 2(a+b+c) = -2\cdot \tfrac{6}{3} = -4$

And, $(b+c)(a+c)+(b+c)(a+b)+(a+c)(a+b) = (a^2+b^2+c^2) + 3(ab+ac+bc)$
This becomes $(a^2+b^2+c^2+2ab+2ac+2bc)+ab+ac+bc = (a+b+c)^2 + (ab+ac+bc) = (-\tfrac{6}{3})^2 - \tfrac{4}{3}$

Finally, $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ac) - abc = (-\tfrac{6}{3})(-\tfrac{4}{3}) - (-\tfrac{7}{3})$

Use that to set up the cubic.

djmccabie · #3 $Old$ October 3rd, 2008, 07:36 AM

Thanks a lot!! you made a bit more work for yourself by repeating most of my working out though but thanks for the last part!!

Soroban · #4 $Old$ October 3rd, 2008, 08:42 AM

Hello, djmccabie!

You did some awesomely excellent work!

Here's one more fact we can dig up . . .

We know: . $\begin{array}{cccc}a + b + c &=& \text{-}2 & {\color{blue}[1]}\\ \\[-4mm] ab+bc + ac &=& \text{-}\frac{4}{3} & {\color{blue}[2]}\\ \\[-4mm] abc &=& \frac{7}{3} & {\color{blue}[3]} \end{array}$

Square [1]: . $(a + b + c)^2 \;=\;(\text{-}2)^2 \quad\Rightarrow\quad a^2 + 2ab + 2ac + b^2 + 2bc + c^2\:=\:4$

$\text{We have: }\;a^2+b^2+c^2 + 2\underbrace{(ab + bc + ac)}_{\text{This is -}\frac{4}{3}} \:=\:4 \quad\Rightarrow\quad a^2 + b^2 + c^2 + 2\left(\text{-}\frac{4}{3}\right) \:=\:4$

Therefore: . $a^2+b^2+c^2\;=\;\frac{20}{3}$

djmccabie · #5 $Old$ October 3rd, 2008, 10:42 AM

just a quick check now. Do you get the answer 3x^3 + 12x^2 + 8x - 15 = 0 ?

(note coefficients must be integers)