Can Mathlab solve this

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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Answer 3

Roger Stafford 2013-9-3

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https://ww2.mathworks.cn/matlabcentral/answers/86245-can-mathlab-solve-this#answer_95737

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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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rob 2013-9-4

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 2013-9-4

编辑：Walter Roberson 2013-9-4

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Suppose you had

a = b^2 + d
b = c^2
c = d^2

then a = d^8 + d, and that has no closed-form solution for d in terms of a. Therefore the generalized 3 x 3 or larger is not always resolvable to algebraic solutions. However, if the forms of the equations are constrained, so that one was not working with the generalized form, then it might be possible to find algebraic solutions; that would vary with the exact constraints.

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