Please help
- Thread starter Farzin
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Please share your work with us ...even if you know it is wrong.
If you are stuck at the beginning tell us and we'll start with the definitions.
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The first thing I would do is to test if any of the unknowns are 0 or 1 because that would make life much easier.
If that does not work, I'd move to the quadratic formula to isolate the unknowns so that I can make substitutions.
\(\displaystyle a^2 + ab + b^2 = 7 \implies a^2 + ab + b^2 - 7 = 0 \implies\)
\(\displaystyle a = \dfrac{-\ b \pm \sqrt{b^2 - 4(1)(b^2 - 7)}}{2} = \dfrac{-\ b \pm \sqrt{28 - 3b^2}}{2}.\)
\(\displaystyle \text {Similarly, } b^2 + bc + c^2 = 28 \implies c = \dfrac{-\ b \pm \sqrt{112 - 3b^2}}{2}.\)
We can now attack the third equation as an equation in one unknown.
\(\displaystyle c^2 + ac + a^2 = 21 \implies\)
\(\displaystyle \left ( \dfrac{-\ b \pm \sqrt{112 - 3b^2}}{2} \right )^2 + \dfrac{-\ b \pm \sqrt{28 - 3b^2}}{2} * \dfrac{-\ b \pm \sqrt{112 - 3b^2}}{2} + \left ( \dfrac{-\ b \pm \sqrt{28 - 3b^2}}{2} \right )^2 = 21.\)
Solving the four equations implied by that mess may be arduous, but it should be possible because they will turn out to be quartics and those are in principle solvable. (If, that is, there are are any real solutions).
EDIT: If you know that the answers are supposed to integers, you can tell that the absolute value of b is small because
\(\displaystyle 3b^2 \le 28 \implies b= -\ 3, -\ 2, -\ 1, 0, 1,\ 2 \text { or } 3.\)
Then you can use trial and error. For example,
\(\displaystyle b = 2 \implies a^2 + 2a + 4 = 7 \implies a^2 + 2a - 3 = 0 \implies\)
\(\displaystyle (a + 3)(a - 1) = 0 \implies a = -\ 3 \text { or } a = 1.\)
\(\displaystyle a = 1 \implies a^2 + ac + c^2 = 21 \implies 1 + c + c^2 = 21 \implies\)
\(\displaystyle c^2 + c - 20 = 0 \implies (c + 5)(c - 4) \implies c = -\ 5 \text { or } c = 4.\)
\(\displaystyle b^2 + bc + c^2 = 4 + 8 + 16 = 28.\)
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mmm4444bot
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I believe there are four solutions. :cool:
Guessing small Integers, one parameter at a time, does help a lot.
I first tried a={-1, 0, 1}
Then I tried b={-1, 0, 1}
Seahawks game in 90 minutes; gotta run! :cool:
Guessing small Integers, one parameter at a time, does help a lot.
I first tried a={-1, 0, 1}
Then I tried b={-1, 0, 1}
Seahawks game in 90 minutes; gotta run! :cool:
This problem has generated a number of questions in my mind, the main one being what is the maximum number of solutions to a system of n equations, none of degree > m. But over at MMF a really elegant short cut was given to the specific question given.
Each of the equations can be restated in terms of a difference of cubes, and from there we can derive a linear equation, which when substituted into two of the quadratics reduces to 4 possible solution sets. It is not a general solution.
Each of the equations can be restated in terms of a difference of cubes, and from there we can derive a linear equation, which when substituted into two of the quadratics reduces to 4 possible solution sets. It is not a general solution.
\(\displaystyle \text {Let } n \text { be the number of unknowns in a system of n independent, consistent p͏olynomial equations.}\)
\(\displaystyle \text {Let } m_j \text { be the highest degree of } x_j \text { in e͏quation } j, \text { with } 1 \le j \le n.\)
In that case, the maximum number of sets of real solutions is \(\displaystyle \displaystyle \prod_{j=1}^n m_j.\)
A perhaps informal proof.
If n = 1, then the maximum number of real solutions is \(\displaystyle m_1\).
Therefore there exists a non-empty set of positive integers, of which k is one, such that the maximum number of solutions is
\(\displaystyle \displaystyle s = \prod_{i=1}^k m_k.\)
Now consider a system of k + 1 equations.
We can find, using the (k + 1)th equation, the (k + 1)th variable in terms of variables 1 through k and then substitute for the (k + 1)th variable in equations 1 through k. That reduced system has at most s solutions.
We can now substitute each of the solutions for the first k variables into the (k + 1)th equation and solve for that last variable. Because the (k + 1)th equation has degree \(\displaystyle m_{k+1}\), it can have as many as \(\displaystyle m_{k+1}\) for each of the solutions for the other variables. Thus, the greatest number of solutions possible is
\(\displaystyle m_{k+1} * s = \displaystyle m_{k+1} * \prod_{i=1}^k m_i = \prod_{i=1}^{k+1}m_i.\)
Thus, in principle, any system of n equations, none of degree greater than 4, has exact solutions determinable by mechanical application of formula.
The given system of 3 quadratic equations with 3 unknowns then could have as many as 8 real solutions. It seems, however, to have only four. Why? The only possible reason is that one of the equations has only one real solution.
\(\displaystyle \text {Let } m_j \text { be the highest degree of } x_j \text { in e͏quation } j, \text { with } 1 \le j \le n.\)
In that case, the maximum number of sets of real solutions is \(\displaystyle \displaystyle \prod_{j=1}^n m_j.\)
A perhaps informal proof.
If n = 1, then the maximum number of real solutions is \(\displaystyle m_1\).
Therefore there exists a non-empty set of positive integers, of which k is one, such that the maximum number of solutions is
\(\displaystyle \displaystyle s = \prod_{i=1}^k m_k.\)
Now consider a system of k + 1 equations.
We can find, using the (k + 1)th equation, the (k + 1)th variable in terms of variables 1 through k and then substitute for the (k + 1)th variable in equations 1 through k. That reduced system has at most s solutions.
We can now substitute each of the solutions for the first k variables into the (k + 1)th equation and solve for that last variable. Because the (k + 1)th equation has degree \(\displaystyle m_{k+1}\), it can have as many as \(\displaystyle m_{k+1}\) for each of the solutions for the other variables. Thus, the greatest number of solutions possible is
\(\displaystyle m_{k+1} * s = \displaystyle m_{k+1} * \prod_{i=1}^k m_i = \prod_{i=1}^{k+1}m_i.\)
Thus, in principle, any system of n equations, none of degree greater than 4, has exact solutions determinable by mechanical application of formula.
The given system of 3 quadratic equations with 3 unknowns then could have as many as 8 real solutions. It seems, however, to have only four. Why? The only possible reason is that one of the equations has only one real solution.
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after solving this, i saw that JeffM gave an insightful and systematic way to solve this, using quadratic formula to find the possibilities for trial and error.
I feel sure this would work for all problems of this nature, and therefore would be a more preferred method, especially if simple trial and error was proving to be too difficult.
Without this insight, I found a=1, b=2 by solving equation one thru pure trial and error.
that left c
I then solved equation 2 and 3 for c2, and set them equal.
28-b2-bc=21-a2-ac
knowing a=1 and b=2, c=4.
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Thank you for your kind words, but, actually, the difficulties of what I proposed soon become very great: you end up trying to solve polynomial equations of the 8th or 9th degree. A system of n equations in n unknowns, where all the equations are at least of degree 2 and at least one equation is of degree m > 2 is truly ugly. You either need tools that are seldom taught in first or second year algebra, or you need information about additional constraints. I tried to work out the general solution for a system of 3 cubic equations in three unknowns and got a severe headache. Even the general problem of two quadratic equations in two unknowns is ugly. It is one thing to show what the maximum number of possible solutions is and quite another thing to find even one of those solutions.
There was an interesting follow-up discussion on this problem, starting at the very bottom of page 1 and continuing onto page 2, at MMF.
http://mymathforum.com/algebra/344655-please-help.html