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A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.). A well-known example is the quadratic formula
The nested radicals in this solution cannot in general be simplified unless the cubic equation has at least one rational solution. Indeed, if the cubic has three irrational but real solutions, we have the casus irreducibilis , in which all three real solutions are written in terms of cube roots of complex numbers.
Algebraic operations in the solution to the quadratic equation.The radical sign √, denoting a square root, is equivalent to exponentiation to the power of 1 / 2 .The ± sign means the equation can be written with either a + or a – sign.
An example of using Newton–Raphson method to solve numerically the equation f(x) = 0. In mathematics, to solve an equation is to find its solutions, which are the values (numbers, functions, sets, etc.) that fulfill the condition stated by the equation, consisting generally of two expressions related by an equals sign.
The quadratic formula =. is a closed form of the solutions to the general quadratic equation + + =. More generally, in the context of polynomial equations, a closed form of a solution is a solution in radicals; that is, a closed-form expression for which the allowed functions are only n th-roots and field operations (+,,, /).
Simplification is the process of replacing a mathematical expression by an equivalent one that is simpler (usually shorter), according to a well-founded ordering. Examples include:
Radical extensions occur naturally when solving polynomial equations in radicals.In fact a solution in radicals is the expression of the solution as an element of a radical series: a polynomial f over a field K is said to be solvable by radicals if there is a splitting field of f over K contained in a radical extension of K.
which if we make the simplifying assumption that b = 0, is equal to + + () This polynomial is of degree six, but only of degree three in z 2, and so the corresponding equation is solvable. By trial we can determine which three roots are the correct ones, and hence find the solutions of the quartic.