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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 ...
In mathematics, a sum of radicals is defined as a finite linear combination of n th roots: =, where , are natural numbers and , are real numbers.. A particular special case arising in computational complexity theory is the square-root sum problem, asking whether it is possible to determine the sign of a sum of square roots, with integer coefficients, in polynomial time.
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.
The square root of a positive integer is the product of the roots of its prime factors, because the square root of a product is the product of the square roots of the factors. Since p 2 k = p k , {\textstyle {\sqrt {p^{2k}}}=p^{k},} only roots of those primes having an odd power in the factorization are necessary.
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
Massachusetts' top court on Friday ruled that a would-be bride must return a $70,000 engagement ring from Tiffany & Co to her former fiancé in a decision that ended 65 years of courts in the New ...
SRS can be solved in polynomial time in the Real RAM model. [3] However, its run-time complexity in the Turing machine model is open, as of 1997. [1] The main difficulty is that, in order to solve the problem, the square-roots should be computed to a high accuracy, which may require a large number of bits.
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