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10-03-2008, 06:28 AM
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| | Further maths. Cubics. What? 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!! | 
10-03-2008, 07:02 AM
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| | Quote:
Originally Posted by djmccabie 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 [x-(b+c)][x - (a+c)] [x - (a+b)] = 0](http://www.mathhelpforum.com/math-help/latex2/img/46fe2000f3f69d61515c3853aecc8d7e-1.gif) .
Note,
And, 
This becomes
Finally,
Use that to set up the cubic.
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10-03-2008, 08:36 AM
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| | 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!! | 
10-03-2008, 09:42 AM
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| | 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} \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}](http://www.mathhelpforum.com/math-help/latex2/img/5c4934bc5a9fe28292d43231328e48df-1.gif)
Square [1]: . 
Therefore: . | | The following users thank Soroban for this useful post: | |  | 
10-03-2008, 11:42 AM
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| | just a quick check now. Do you get the answer 3x^3 + 12x^2 + 8x - 15 = 0 ?
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