Search results
Results from the WOW.Com Content Network
The Babylonian clay tablet YBC 7289 (c. 1800–1600 BC) gives an approximation of the square root of 2 in four sexagesimal figures, 𒐕 𒌋𒌋𒐼 𒐐𒐕 𒌋 = 1;24,51,10, [13] which is accurate to about six decimal digits, [14] and is the closest possible three-place sexagesimal representation of √ 2:
Plimpton 322 is a Babylonian clay tablet, believed to have been written around 1800 BC, that contains a mathematical table written in cuneiform script.Each row of the table relates to a Pythagorean triple, that is, a triple of integers (,,) that satisfies the Pythagorean theorem, + =, the rule that equates the sum of the squares of the legs of a right triangle to the square of the hypotenuse.
The solutions of the quadratic equation ax 2 + bx + c = 0 correspond to the roots of the function f(x) = ax 2 + bx + c, since they are the values of x for which f(x) = 0. If a , b , and c are real numbers and the domain of f is the set of real numbers, then the roots of f are exactly the x - coordinates of the points where the graph touches the ...
Babylonian clay tablet YBC 7289 with annotations. The diagonal displays an approximation of the square root of 2 in four sexagesimal figures, 1 24 51 10, which is good to about six decimal digits. 1 + 24/60 + 51/60 2 + 10/60 3 = 1.41421296... The tablet also gives an example where one side of the square is 30, and the resulting diagonal is 42 ...
It is also used for graphing quadratic functions, deriving the quadratic formula, and more generally in computations involving quadratic polynomials, for example in calculus evaluating Gaussian integrals with a linear term in the exponent, [2] and finding Laplace transforms. [3] [4]
The quadratic equation on a number can be solved using the well-known quadratic formula, which can be derived by completing the square. That formula always gives the roots of the quadratic equation, but the solutions are expressed in a form that often involves a quadratic irrational number, which is an algebraic fraction that can be evaluated ...
A similar but more complicated method works for cubic equations, which have three resolvents and a quadratic equation (the "resolving polynomial") relating and , which one can solve by the quadratic equation, and similarly for a quartic equation (degree 4), whose resolving polynomial is a cubic, which can in turn be solved. [14]
In stage 2, the well-attested Old Babylonian method of completing the square is used to solve what is effectively the system of equations b − a = 0.25, ab = 0.75. [6] Geometrically this is the problem of computing the lengths of the sides of a rectangle whose area A and side-length difference b − a are known, which was a recurring problem ...