whats the defintio for solving equation?
30950
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572493
=x(x^3-2)\n--> x=0 or x=2^(1/3) or -2^(1/3)
429670
In matlab, type\n>> syms x\n>> solve('x^3+4*x^2-5=0')\n \nans =\n \n[                1]\n[ -5/2+1/2*5^(1/2)]\n[ -5/2-1/2*5^(1/2)]
532355
Please see the method explained by me as an answer tothis question..\nhttp://in.answers.yahoo.com/question/index;_ylt=AiNJtRa.xTUMUTdP29hR7sSQHQx.?qid=1006051102360\n\nThis will help u solve cubic as well as other equations.. :-)
660658
Let y = f(x) be the solution with y(0) = 1, y'(0) = 0.\n\nLet y = g(x) be the solution with y(0) = 0, y'(0) = 1.\n\nThen the solution for y(0) = c0, y(1) = c1 is just:\ny(x) = c0 f(x) + c1 g(x)\n\nAs power series, we can write:\n\nf(x) = 1 + 0 x + a2 x^2 + a3 x^3 + a4 x^4 + ...\n\ng(x) = 0 + 1 x + b2 x^2 + b3 x^3 + b4 x^4 + ...\n\nThen f''(x) = 2 a2 + 6 a3 x + 12 a4 x^2 + ...\nwith (n+2) (n+1) a{n} as the coefficient of x^n.\n\nLikweise for g''(x) but replacing the a's with b's.\n\nFinally, sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - ...\nwith all x^even coefficients equal zero, and\nwith the x^(2k+1) coefficient equal to (-1)^k / (2k+1)!.\n\nNow just express both\n\nxf'' + sin(x) f = 0\nxg'' + sin(x) g = 0\n\nas power series on the left (messy) and right (all zeroes)\nand equate left and right (zero) coefficients for each of\nx^0, x^1, x^2, x^3, etc.\n\nFor f you have\n\n(2 a2 x + 6 a3 x^2 + 12 a4 x^3 + ...) +\n (x - (1/3!)x^3 + (1/5!)x^5 - ...) (1 + x + a2 x^2 + a3 x^3 + ...)\n = 0\n\n0 x^0\n + (2 a2 + 1) x^1\n + (6 a3 + 1) x^2\n + (12 a4 + a2 - (1/6)) x^3\n + ...\n = 0 = 0 + 0 x + 0 x^2 + 0 x^3 + ...\n\nEquating coefficients, you get\n\n0 = 0\n that is good!\n\n2 a2 + 1 = 0\n or a2 = -1/2\n\n6 a3 + 1 = 0\n or a3 = -1/6 \n\n12 a4 + a2 - (1/6) = 0\n or 12 a4 = (1/6) - (-1/2) = 2/3\n or a4 = 1/18\n\nSo f(x) = 1 + (-1/2) x^2 + (-1/6) x^3 + (1/18) x^4 + ...\n\nFor g it is similar except that you start with:\n\n(2 b2 x + 6 b3 x^2 + 12 b4 x^3 + ...) +\n (x - (1/3!)x^3 + (1/5!)x^5 - ...) (x + b2 x^2 + b3 x^3 + ...)\n = 0
797913
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
354066
you need to give more information, solve for what? this isn't an equation, there is no equals sign so nothing to solve. Its just a statement
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
920401
usually when you solve equations you get a solution.\nthe solution solves your problem.\n!
886265
gotta put the equation into y=mx+b first\ny= 1/3x - 1\n\nperpendicular would be y = -3x (+ or - any #)
764068
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
169494
The answer is:\n2y^3 - y^2\nOR\n(2y-1)*y^2
750522
Begin with where you left of:\n\n2x + 2yy' = 0\n\nDifferentiate again, just like the first time.  Use the chain rule on the yy' term:\n\n2 + 2(y'y' + yy'') = 0\n1 + y'^2 + yy'' = 0\ny'' = (-1-y'^2)/ y\n\nFrom your y' = -x/y\n\ny'' = (-1-(-x/y)^2)/ y = -x^2/y^3 - 1/y\n\nThis isn't the answer you have but see if they are the same:\n\nx^2 = 4 - y^2\n\ny'' = -(4-y^2)/y^3 - 1/y = -4/y^3 + 1/y - 1/y = -4/y^3\n\nSo the two solutions are the same
234829
With factorisation you turn the equation x^2 + px + q = 0 into two smaller equations in brackets like this:\n\n(x+a)(x+b) = 0\n\nSo that a+b = p\nand a*b = q\n\nEG. Your first example you have p = 4 and q = 3\n\nSo you've got a+b = 4 so a could be 2 and b 2 or a 1 and b 3 etc\n\nand a*b = 3 so a could be 1 and b 3... Yes it matches.\n\n(x+3)(x+1) = 0\n\nThe solution is just x = -a and x = -b\n\n-> x = -1 or x = -3
531205
2 + 2 = 4\n\nthats all I got
117786
Well, there isn't really anything to "solve", since it's not an equation.  If you want to simplify, it depends what your original terms are.  Is the question\n\na)  2x^2 + 4/2x  ?\nIf so, then the only simplification is 2x^2 + 2/x.\n\nb) (2x^2 + 4) / 2x ?\nIn this case, the simplification is (x^2 + 2) / x.\n\nI hope this answers your question!  Good luck.