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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 ...
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.
In this vein, the discriminant is a symmetric function in the roots that reflects properties of the roots – it is zero if and only if the polynomial has a multiple root, and for quadratic and cubic polynomials it is positive if and only if all roots are real and distinct, and negative if and only if there is a pair of distinct complex ...
Fermat's method works best when there is a factor near the square-root of N. If the approximate ratio of two factors ( d / c {\displaystyle d/c} ) is known, then a rational number v / u {\displaystyle v/u} can be picked near that value.
This corresponds to a set of y values whose product is a square number, i.e. one whose factorization has only even exponents. The products of x and y values together form a congruence of squares. This is a classic system of linear equations problem, and can be efficiently solved using Gaussian elimination as soon as the number of rows exceeds ...
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 fourth section derives approximations, both numerical and geometrical, of irrational numbers such as square roots. [10] The book also includes proofs in Euclidean geometry. [11] Fibonacci's method of solving algebraic equations shows the influence of the early 10th-century Egyptian mathematician Abū Kāmil Shujāʿ ibn Aslam. [12]
We are not taking the square root of any negative values here, since both and are necessarily positive. But we have lost the solution x = − 2. {\displaystyle x=-2.} The reason is that x {\displaystyle x} is actually not in general the positive square root of x 2 . {\displaystyle x^{2}.}