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However, more insidious are missing solutions, which can occur when performing operations on expressions that are invalid for certain values of those expressions. For example, if we were solving the following equation, the correct solution is obtained by subtracting from both sides, then dividing both sides by :
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 process of simplifying expressions involving the square root of an expression involving the square root of another expression involves finding the two solutions of a quadratic equation. Descartes' theorem states that for every four kissing (mutually tangent) circles, their radii satisfy a particular quadratic equation.
The solution set for the equations = and + = is the single point (2, 3). An example of solving a system of linear equations is by using the elimination method: {+ = = Multiplying the terms in the second equation by 2:
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 solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation. If all of the coefficients a , b , c , and d of the cubic equation are real numbers , then it has at least one real root (this is true for all odd-degree polynomial functions ).
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 (+,,, /).
The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0, a mathematical truth. But the same substitution applied to the original equation results in x/6 + 0/0 = 1, which is mathematically meaningless.
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