Who discovered  the cubic equation and the formula for the solution??
930926
- The solution to the cubic (as well as the quartic) was published by Gerolamo Cardano (1501-1576) in his treatise Ars Magna. However, Cardano was not the original discoverer of either of these results. The hint for the cubic had been provided by Niccolò Tartaglia, while the quartic had been solved by Ludovico Ferrari. However, Tartaglia himself had probably caught wind of the solution from another source. The solution was apparently first arrived at by a little-remembered professor of mathematics at the University of Bologna by the name of Scipione del Ferro (ca. 1465-1526). While del Ferro did not publish his solution, he disclosed it to his student Antonio Maria Fior. This is apparently where Tartaglia  learned of the solution around 1541.
498372
I think u are talkin abt this method....\n\n\nif u hav a cubic like  x^3 + 2x^2 - 2x - 1 = 0\n\nto find out whether (x-1) is a factor....\n\nfirst write the coeff of the eq in this way n write the factor ( here it iz 1 since x-1=0 or x=1) on the left hand side Like this\n\n{one}  1     2     -2     -1\n--------  0     1      3      1\n_______________________________\n{sum}  1     3      1      0\n\n\nHere what i hav done is write the 0 under the coeff of x^3\nnow add that coeff n the zero\nu get 1...\nnow multiply this one wid the number written on the left hand side(here it is also one)\nwite what u get under the coeff of x^2....\nas above add the new no and the coeff of x^2\nin a similar way multiply one and the sum u got n write it under the coeff of x...\nadd the no with coeff of x n then write it below....\nfinally u multiply the sum wid one and write it under the constant term.... n then add...\nif u get 0 as the sum then whatever no u hav taken in da LHS is on of the root.... here we get 0 in the last case... so (x-1) is a fctor.....\n\nnow this eq can b written as \n\n    (x-1) * (1*x^2 + 3 * x + 1) = 0\nhere 1 , 3 and 1 of the quadratic are the sum u got in the above method......
55082
No no solution since n>2\n\nPierre de Fermat died in 1665. Today we think of Fermat as a number theorist, in fact as perhaps the most famous number theorist who ever lived. It is therefore surprising to find that Fermat was in fact a lawyer and only an amateur mathematician. Also surprising is the fact that he published only one mathematical paper in his life, and that was an anonymous article written as an appendix to a colleague's book.\n\nThere is a statue of Fermat and his muse in his home town of Toulouse:\n(Click it to see a larger version)\n\n\nBecause Fermat refused to publish his work, his friends feared that it would soon be forgotten unless something was done about it. His son, Samuel undertook the task of collecting Fermat's letters and other mathematical papers, comments written in books, etc. with the object of publishing his father's mathematical ideas. In this way the famous 'Last theorem' came to be published. It was found by Samuel written as a marginal note in his father's copy of Diophantus's Arithmetica. \n\nFermat's Last Theorem states that \n\nxn + yn = zn \n\nhas no non-zero integer solutions for x, y and z when n > 2. Fermat wrote \n\nI have discovered a truly remarkable proof which this margin is too small to contain.
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
831017
sir isaac newton. the man who "discovered" the law of gravity.
145305
Use cubic equation \ny=ax3+bx2+cx+d\n\nSince you have four points data you can get four linear equations with four variables which is very easy to solve, get variables (a,b,c,d) and plug the values to cross check and see u will get answer. Use guass jordan or elimination method.
1017464
It is the equation for cubic equations : \n\n             \nx=cube root[(-q/2)+sqrt((q/2)^2+\n(p/3)^3)]+\ncube root[(-q/2)-sqrt((q/2)^2+\n(p/3)^3)]\n\nIt's definetly the most beautiful, if not the most complex.
994889
Start with a general cubic equation:\nx^3 +ax^2 +bx+c=0.\nFirst, we want to get rid of the x^2 term, so let \nx=y-(a/3).\nIf you plug this in and expand, the y^2 terms will drop out and you will get an equation of the form\ny^3 +Cy+D=0.\nNow, if you let y=u+v (to get an extra bit of freedom), and expand, you get\nu^3 +v^3 +(3uv+C)(u+v)+D=0.\nChoose v so that 3uv+C=0, i.e. v=-C/(3u). Plug this into\nu^3 +v^3 +D=0\nand multiply through by u^3 to get a quadratic in u^3. Solve\nthis by the quadratic formula and take a cube root to find\nu. There will be three cube roots in the complex plane, so find all of them. For each u, find v=-C/(3u), then y=u+v, and finally x=y-(a/3).\n\nThis is often a rather cumbersome proceedure, even if you do some of it ahead of time. Also, the roots may not 'look' like what you expect. If you apply this to an equation that has a root of, say, 4, you will obtain 4 as a sum of two cube roots of quadratic expressions. To recognize the answer as 4 is a challenge in and of itself.
913819
The "Hindu Solution" to the quadratic equation is the same method that we now call "completing the square."  This is a way to convert terms in the equation to the form of the square of a different polynomial, enabling it to be solved algebraically.  When applied to a quadratic equation in the general form, this method yields the quadratic formula.
768295
Using an equation is the lazy, mechanical way to do it. If you want to understand what's going on, factor it if possible.\n\nIf you *really* want to know what's going on, derive the quadratic formula yourself. It isn't as hard as it looks, and it can be done on one page.
946795
Carl Gauss, the great scientist and mathematician (1777--1855). He invented it, when during a class, his teacher wanted to quiet down the pupils with "busy work", by having them add up all of the numbers between 1--100. Normally, adding it by the method of 1+2+3.... would've taken quite a long time. Gauss did it in his head, by adding the composite numbers from the "outside in" -- 1+100, 2+99, 3+98 etc (101 x 50). He solved the problem in a matter of minutes.
443349
We are given:\nx^1/2 + y = 11 ---- eqn. 1\ny^1/2 + x = 7 ----- eqn. 2\n\nLet a = x^1/2. Therefore x = a^2\nLet b = y^1/2. Therefore, y = b^2\n\na + b^2 = 11 ==== eqn. 3\na^2 + b = 7 ==== eqn. 4\n\nSub. (3) into eqn. (4):\n\n=> (11 - b^2)^2 + b = 7\n=> 121 - 22b^2 + b^4 + b = 7\n=> b^4 - 22b^2 + b + 114 = 0\n\nUsing trial and error, we can arrive that (b-3) is a factor.\nTherefore, by factorisation:\n=> b^4 - 22b^2 + b + 114 = 0\n=> (b-3) (b^3 +3b^2 - 13b - 38)=0\nTherefore, b = 3  (there are no other possible root values for b because the cubic equation lies completely above the x-axis.)\n\nSince b = 3, therefore, y = 9. And back into eqn. 1, x= 4.
863763
D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
1174830
I don't think he asked for a solution of an eqn. but for factors\n\nx^4 + y^4 = (x^2 + y^2)^2  - 2.x^2.y^2\n\nRHS can be expressed as difference of 2 squares\nAnd so on.  I don't feel like doing the rest. you get the idea.
218913
The formula is credited to Heron of Alexandria in the 1st century, and a proof can be found in his book Metrica. It is now believed that Archimedes already knew the formula, and it is of course possible that it has been known long before.\n\nLet the sides of a triangle have lengths a,b and c. Introduce the semiperimeter p = (a + b + c)/2 and the area S. Then Heron's formula asserts that\n S2 = p(p - a)(p - b)(p - c)\n\nW.Dunham analyzes the original Heron's proof in his Journey through Genius.\n\nFor the right triangle with hypotenuse c, we have S = ab/2. We'll modify the right hand side of the formula by noting that\n p - a = (- a + b + c)/2, p - b = (a - b + c)/2, p - c = (a + b - c)/2\n\nIt takes a little algebra to show that\n \n(a + b + c)(- a + b + c)(a - b + c)(a + b - c)\n= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)\n\nFor the right triangle, this expression is equal to 16S2 = 4a2b2. So we have\n 4a2b2= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)\n\nTaking all terms to the left side and grouping them yields\n (a4 + 2a2b2 + b4) - 2a2c2 - 2b2c2 + c4 = 0\n\nWith a little more effort\n (a2 + b2)2 - 2c2(a2 + b2) + c4 = 0\n\nAnd finally\n [(a2 + b2) - c2]2 = 0\nRemark\n\nFor a quadrilateral with sides a, b, c and d inscribed in a circle there exists a generalization of Heron's formula discovered by Brahmagupta. In this case, the semiperimeter is defined as p = (a + b + c + d)/2. Then the following formula holds\n S2 = (p - a)(p - b)(p - c)(p - d)\n\nSince any triangle is inscribable in a circle, we may let one side, say d, shrink to 0. This leads to Heron's formula.\n\nFor more information you can go to www.wikipedia.com  \nhttp://mathworld.wolfram.com/HeronsFormula.html\nhttp://www.cut-the-knot.org/pythagoras/herons.shtml\n\nI hope this helps! =-)
792339
A quadratic equation is a second-order polynomial equation in a single variable x\nax^2+bx+c==0,\n(1)\n\nwith a!=0. Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions. These solutions may be both real, or both complex.\n\nAmong his many other talents, Major General Stanley in Gilbert and Sullivan's operetta the Pirates of Penzance impresses the pirates with his knowledge of quadratic equations in "The Major General's Song" as follows: "I am the very model of a modern Major-General, I've information vegetable, animal, and mineral, I know the kings of England, and I quote the fights historical, From Marathon to Waterloo, in order categorical; I'm very well acquainted too with matters mathematical, I understand equations, both the simple and quadratical, About binomial theorem I'm teeming with a lot o' news-- With many cheerful facts about the square of the hypotenuse."\n\nThe roots x can be found by completing the square,\nx^2+b/ax==-c/a\n(2)\n(x+b/(2a))^2==-c/a+(b^2)/(4a^2)==(b^2-4ac)/(4a^2)\n(3)\nx+b/(2a)==(+/-sqrt(b^2-4ac))/(2a).\n(4)\n\nSolving for x then gives\nx==(-b+/-sqrt(b^2-4ac))/(2a).\n(5)\n\nThis equation is known as the quadratic formula.\n\nThe first known solution of a quadratic equation is the one given in the Berlin papyrus from the Middle Kingdom (ca. 2160-1700 BC) in Egypt. This problem reduces to solving\nx^2+y^2=100\n(6)\ny=3/4x\n(7)\n\n(Smith 1953, p. 443). The Greeks were able to solve the quadratic equation by geometric methods, and Euclid's Eric Weisstein's World of Biography (ca. 325-270 BC) Data contains three problems involving quadratics. In his work Arithmetica, the Greek mathematician Diophantus Eric Weisstein's World of Biography (ca. 210-290) solved the quadratic equation, but giving only one root, even when both roots were positive (Smith 1951, p. 134).\n\nA number of Indian mathematicians gave rules equivalent to the quadratic formula. It is possible that certain altar constructions dating from ca. 500 BC represent solutions of the equation, but even should this be the case, there is no record of the method of solution (Smith 1953, p. 444). The Hindu mathematician Aryabhata (475 or 476-550) gave a rule for the sum of a geometric series that shows knowledge of the quadratic equations with both solutions (Smith 1951, p. 159; Smith 1953, p. 444), while Brahmagupta Eric Weisstein's World of Biography (ca. 628) appears to have considered only one of them (Smith 1951, p. 159; Smith 1953, pp. 444-445). Similarly, Mahavira (ca. 850) had substantially the modern rule for the positive root of a quadratic. Sridhara (ca. 1025) gave the positive root of the quadratic formula, as stated by Bhaskara (ca. 1150; Smith 1953, pp. 445-446). The Persian mathematicians al-Khwarizmi Eric Weisstein's World of Biography (ca. 825) and Omar Khayyám Eric Weisstein's World of Biography (ca. 1100) also gave rules for finding the positive root.\n\nViète Eric Weisstein's World of Biography was among the first to replace geometric methods of solution with analytic ones, although he apparently did not grasp the idea of a general quadratic equation (Smith 1953, pp. 449-450).\n\nAn alternate form of the quadratic equation is given by dividing (◇) through by x^2:\na+b/x+c/(x^2)==0\n(8)\nc(1/(x^2)+b/(cx))+a==0\n(9)\nc(1/x+b/(2c))^2==c(b/(2c))^2-a==(b^2)/(4c)-(4ac)/(4c)==(b^2-4ac)/(4c).\n(10)\n\nTherefore,\n1/x+b/(2c)==+/-(sqrt(b^2-4ac))/(2c)\n(11)\n1/x==(-b+/-sqrt(b^2-4ac))/(2c)\n(12)\nx==(2c)/(-b+/-sqrt(b^2-4ac)).
706946
In general it is difficult to solve a cubic equation; there is an algebraic formula for the solutions of a general cubic, but it is extremely complicated.\n\nTry to find a solution by inspection. First I will use the rational-roots test, which is that all rational roots of a polynomial (if they exist) can be written in the form a/b, where a divides the constant term and b divides the leading coefficient. This means that the only possible rational roots are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, and their negatives. It is not hard to see that none of these is a root.\n\nNow I will try elements of the extension fields of the rationals by square roots of factors of 60.\n\nIf x=a+b√2, then a^3+3a^2*b√2+6ab^2+2b^3√2+2a+2b√2=60, so a^3+6ab^2+2a=60, and 3a^2*b+2b^3+2b=0, so 3a^2+2b^2=-2, which is impossible if a and b are rational; similar things happen with any other extension fields of degree 2.\n\nYou can also try extension fields of higher degree to see whether you get lucky.\n\nAnother idea is to find two values of x such that x^3+2x-60 is negative for one value and positive for another (for example, 3 and 4), and use procedures like Newton's method or regula falsi to approximate the real root (a graph indicates that there is only one real root).\nI cheated with my calculator and found no easy factorization.