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In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
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
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 unresolved root, especially one using the radical symbol, is sometimes referred to as a surd [2] or a radical. [3] Any expression containing a radical, whether it is a square root, a cube root, or a higher root, is called a radical expression , and if it contains no transcendental functions or transcendental numbers it is called an algebraic ...
A quadratic equation is one which includes a term with an exponent of 2, for example, , [40] and no term with higher exponent. The name derives from the Latin quadrus , meaning square. [ 41 ] In general, a quadratic equation can be expressed in the form a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0} , [ 42 ] where a is not zero (if it were ...
A method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with ...
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.
In number theory, the radical of a positive integer n is defined as the product of the distinct prime numbers dividing n. Each prime factor of n occurs exactly once as a factor of this product: r a d ( n ) = ∏ p ∣ n p prime p {\displaystyle \displaystyle \mathrm {rad} (n)=\prod _{\scriptstyle p\mid n \atop p{\text{ prime}}}p}