Root Finding Methods!  Newtons! Bisectional!  Algebra!?
922218
I found the Wikipedia's article on Newton's method to be very good:\n\nhttp://en.wikipedia.org/wiki/Newton's_method
791450
assuming h is a real number, we have:\nto find roots of any function set it equal to zero\n 1 - tan(h)*[2*pi*h/x]^2 = 0\n      tan(h)*[2*pi*h/x]^2 = 1 \ntan(h)*[4*pi^2*h^2/x^2] = 1\ntan(h)*[4*pi^2*h^2] = x^2  x ~= 0 (not equal to zero)\nx = +/- sqrt(tan(h))*2*pi*h these are the roots
868180
the roots are (2a-1)x(2a-1)
67339
See the explanation in the Wikipedia here:\n\nhttp://en.wikipedia.org/wiki/Methods_of_computing_square_roots\n\nScroll down the the section marked "An exact long-division-like algorithm".
358881
http://www.nist.gov/dads/HTML/squareRoot.html\nhttp://www.math.com/students/calculators/source/square-root.htm\n\nI hope it will help
978661
Have fun:\nhttp://www.math.nyu.edu/~griffith/teaching/06numanalysis/labs/bisection/
264348
The best way to start is to create a function based on the equation. Do this by setting the equation equal to 0. For instance, the first step to solving the equation x^3=5 would be to subtract 5 from both sides to get x^3-5=0. This creates the function f(x)=x^3-5.\n\nNow, look at the graph of the function. Are there any obvious places it appears to cross the x-axis? If the function appears to cross at x = 2 then perform synthetic division, dividing the cubic equation by x-2 (in other words, put a +2 in the synthetic division "box"). If you get a remainder, then it wasn't a root. If there's no remainder, you've factored out the root, and you'll get (x-2)(ax^2+bx+c), where a, b, and c are the numbers that come out of the synthetic division you just completed.\n\nNow you can factor ax^2+bx+c (if it's factorable) or use the quadratic formula (if it's not) to get the other two solutions. There will be 3, even if some of them are imaginary (have negatives inside radicals).\n\nIf none of the x-intercepts are rational (none of the synthetic division worked), pick an x-value close to each x-intercept on the graph and apply Newton's method to each one to estimate each root as accurate as you'd like.\n\nClick on the link below to see how I apply Newton's Method to solve f(x) = x^3-5. It's a big graphic, and your browser may smoosh it to make it fit in the screen, which makes it unreadable. Usually, clicking on the image or holding your mouse over the image until a resize box appears in the lower right corner will let you expand it to the real size. Either that or save it to your computer and double-click it there.\n\nBest of luck.
910180
3ab(2ab^2)^4\n3ab(16a^4b^8)\n48a^5b^9\ntrust me!!!! i just got done doing this in class a week ago.
82007
Just type in quadratic equation and maybe roots in a serch engine. You will have your pick of dozens of sites.
1127430
try this one http://www.algebrahelp.com/
832512
either square root or erauqs
789094
divide by x^2\n6(x^2+(1/x^2))+ (x)+(1/x)+10 = 0\nput x+(1/x)= t\n\nthen solve for t and then solve for x\nyou'll find that there are no roots.!
964615
Please type smith's algorithm for polynomial matrices at yahoo search you get links hope fully one of them should provide you with a solution.
919864
1.  ax + xb\n2.  21x^3 + 24x*2y\n3.  24x^2 + 31x + 10\n4.  24x^2 - 8x - 6\n5.  40x^3 + 66x^2 + 38x + 6\n6.  x^2 - 1\n7.  x^3 - 8
520168
Try this:\n\nhttp://www.algebra.com/
1115504
the cube root can be found using a calculator\nmake sure you have a scientific one though\ni had do use it alot in geometry\n\nits alot like a square root just three instead of 2
826907
ill write down the formula for finding the root of a quadratic equation :  ax^2 + bx + c = 0\nsolution: x = {-b +- sqrt(b^2 - 4ac) } / 2a\n\nequation x^2=5x + 2 can be written as :\n\nx^2 - 5x - 2 = 0\nhere a = 1 ; b = -5 ; c = -2 ;\nsimply apply the formula to find the value of x .\n\nx ={ -(-5)+-sqrt(25 - 4*-2) } /  2\nx ={ 5 +- sqrt(24+8) } / 2\nx = { 5 +- sqrt(33) } / 2\n\nThe second equation :\n\n4x^2 + 2x - 3 = 0\nhere a = 4 ; b = 2 ; c = -3\n\nx = {-2 +- sqrt( 4 - 4*4*-3) } / 2*4\nx = {-2 +- sqrt(4+48) } / 8\nx = { -2 +- sqrt(52) } /8\n\nsimplify the term inside the square root ,  52 = 13*4\nso 4 gets square rooted and\n\nx = { -2 +- 2 sqrt(13) } /8\n\ntake 2 common from the numerator and denominator and cancel them.\n\nx = 2{ -1 +- sqrt(13) } / 4*2\nx = { -1 +- sqrt(13) } / 4\n\nthats your answer.\n\nhave a nice day :)