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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
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 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 ...
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. The problem is mentioned in the Open Problems Garden. [4] Blomer [5] presents a polynomial-time Monte Carlo algorithm for deciding whether a
A root of degree 2 is called a square root and a root of degree 3, a cube root. Roots of higher degree are referred by using ordinal numbers, as in fourth root, twentieth root, etc. The computation of an n th root is a root extraction. For example, 3 is a square root of 9, since 3 2 = 9, and −3 is also a square root of 9, since (−3) 2 = 9.
The solution = is in fact a valid solution to the original equation; but the other solution, =, has disappeared. The problem is that we divided both sides by x {\displaystyle x} , which involves the indeterminate operation of dividing by zero when x = 0. {\displaystyle x=0.}
Newton's method is a powerful technique—if the derivative of the function at the root is nonzero, then the convergence is at least quadratic: as the method converges on the root, the difference between the root and the approximation is squared (the number of accurate digits roughly doubles) at each step. However, there are some difficulties ...
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