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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.
Let () be a polynomial equation, where P is a univariate polynomial of degree n. If one divides all coefficients of P by its leading coefficient c n , {\displaystyle c_{n},} one obtains a new polynomial equation that has the same solutions and consists to equate to zero a monic polynomial.
In mathematics, a rational function is any function that can be defined by a rational fraction, which is an algebraic fraction such that both the numerator and the denominator are polynomials. The coefficients of the polynomials need not be rational numbers ; they may be taken in any field K .
A rational algebraic expression (or rational expression) is an algebraic expression that can be written as a quotient of polynomials, such as x 2 + 4x + 4. An irrational algebraic expression is one that is not rational, such as √ x + 4.
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.
In mathematics, specifically in elementary arithmetic and elementary algebra, given an equation between two fractions or rational expressions, one can cross-multiply to simplify the equation or determine the value of a variable.
In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.
This rational number can be found by realizing that x also appears under the radical sign, which gives the equation x = 2 + x . {\displaystyle x={\sqrt {2+x}}.} If we solve this equation, we find that x = 2 (the second solution x = −1 doesn't apply, under the convention that the positive square root is meant).