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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.
An algebraic number is a number that is a root of a non-zero polynomial in one variable with integer (or, equivalently, rational) coefficients. For example, the golden ratio, (+) /, is an algebraic number, because it is a root of the polynomial x 2 − x − 1. That is, it is a value for x for which the polynomial evaluates to zero.
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.}
The square root of 2 is equal to the length of the hypotenuse of a right triangle with legs of length 1 and is therefore a constructible number. In geometry and algebra, a real number is constructible if and only if, given a line segment of unit length, a line segment of length | | can be constructed with compass and straightedge in a finite number of steps.
The modular square root of can be taken this way. Having solved the associated quadratic equation we now have the variables w and set v = r (if C in the quadratic is a natural square). Solve for variables α {\displaystyle \alpha } and β {\displaystyle \beta } the following equation:
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By performing this iteration, it is possible to evaluate a square root to any desired accuracy by only using the basic arithmetic operations. The following three tables show examples of the result of this computation for finding the square root of 612, with the iteration initialized at the values of 1, 10, and −20.