# Can Mathlab solve this

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rob 3 Sep 2013
Can Mathlab solve this
x1^2 +2.x1 - 2.x2^2 -5.x2 =5
2.x1^2 -3.x1 +x2^2 +3.x2 =19

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Walter Roberson 3 Sep 2013
You know that has four solutions, right?
Walter Roberson 3 Sep 2013

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### 답변(3개)

Thomas 3 Sep 2013
편집: Thomas 3 Sep 2013
Yes, look in the symbolic math toolbox http://www.mathworks.com/help/symbolic/solve.html
Go to the bottom of the page for examples

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Shashank Prasanna 3 Sep 2013
You can solve a system of nonlinear equations using FSOLVE:
This will yield numerical solutions for x1 and x2

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Walter Roberson 3 Sep 2013
Note: fsolve() will only find one solution at a time.
rob 3 Sep 2013
fsolve is all numerical not algabraic
Walter Roberson 3 Sep 2013
Correct, fsolve() is numeric not algebraic. However can you really make use of the algebraic solutions? For example one of the four solutions to the above system has x1 be
-349/140 + (1/5040) * (4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3) - (1/90720) * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344)/(112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) * 6^(1/2) * 36^(1/2) * 2^(1/2) * (((810* (112706532 + 2940 * 1239703701^(1/2))^(1/3) + 18^(1/3) * ((9392211 + 245 * 1239703701^(1/2))^2)^(1/3) + 62862) * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) + 1764 * (112706532 + 2940 * 1239703701^(1/2))^(1/3)) * 18^(1/3) * ((112706532 + 2940 * 1239703701^(1/2))^(1/3) / (810 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 125724))^(1/2) * ((9392211 + 245 * 1239703701^(1/2))^2)^(1/3) * 6^(1/2) / (9392211 + 245 * 1239703701^(1/2)))^(1/2) + (1/30240) * (1620 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) + 2 * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) * 18^(1/3) * ((9392211 + 245 * 1239703701^(1/2))^2)^(1/3) + 125724 * ((4860 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - 6 * (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 754344) / (112706532 + 2940 * 1239703701^(1/2))^(1/3))^(1/2) + 3528 * (112706532 + 2940 * 1239703701^(1/2))^(1/3)) * 6^(1/2) * ((112706532 + 2940 * 1239703701^(1/2))^(1/3) / (810 * (112706532 + 2940 * 1239703701^(1/2))^(1/3) - (112706532 + 2940 * 1239703701^(1/2))^(2/3) - 125724))^(1/2) * 18^(1/3) * ((9392211 + 245 * 1239703701^(1/2))^2)^(1/3) / (9392211 + 245 * 1239703701^(1/2))
Would your work seriously be affected if all those 112706532 where 112706533 instead? (That would make a difference in the 6th decimal place.)

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Roger Stafford 3 Sep 2013
It is useful to know how to solve such equations by hand rather than always depending on matlab. The trick is to eliminate either the x1^2 term or the x2^2 term by combining the equations appropriately. If we double the second equation and then add the equations, we get
5*x1^2-4*x1+x2 = 43
which can be solved for x2
x2 = -5*x1^2+4*x1+43
You can then substitute this value of x2 into either one of the original equations and get a fourth degree polynomial equation in x1. The four roots of this can be obtained with matlab's 'roots' program (we need matlab after all) and then corresponding values of x2 from these with the above equation.

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Walter Roberson 3 Sep 2013
You have not defined "it" sufficiently for an answer.
In your example, each variable appears only in the first or second degree, and never in the third degree or higher, and never any fractional or negative exponent. The constant multipliers and the value equated to are all rational and integer in your example. Your example also has no cross-products, no occurrence of x1^n * x2^m for n, m > 0. Effectively, each of your entries is the sum of exactly two quadratic equations with integer coefficients.
in the 3 x 3 and larger versions, which (if any) of those conditions is modified? Is each entry in the 1000 x 1000 still the sum of exactly two quadratic equations in integer coefficients?
rob 4 Sep 2013
I dont understand exactly I think lineair dependence is solvable like a +b +c =3 2a +2b - c =5 -a +b -3c = 9
but i was investigating only where the a b and c have a quadratic So a thousend by a thousend has a thousend unkowns of a and a^2 for all i know this is unsolvable and only in fsolve with numerical math.
Walter Roberson 4 Sep 2013