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The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
In the mathematical discipline of numerical linear algebra, a matrix splitting is an expression which represents a given matrix as a sum or difference of matrices. Many iterative methods (for example, for systems of differential equations) depend upon the direct solution of matrix equations involving matrices more general than tridiagonal matrices.
The solution set for the equations x − y = −1 and 3x + y = 9 is the single point (2, 3). A solution of a linear system is an assignment of values to the variables ,, …, such that each of the equations is satisfied. The set of all possible solutions is called the solution set. [5]
For example, the system x 3 – 1 = 0, x 2 – 1 = 0 is overdetermined (having two equations but only one unknown), but it is not inconsistent since it has the solution x = 1. A system is underdetermined if the number of equations is lower than the number of the variables.
The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯ are 1, 3, 6, 10, 15, etc.The nth partial sum is given by a simple formula: = = (+). This equation was known ...
A necessary (but not sufficient) condition for solvability is that n is not divisible by 4 or by a prime of form 4k + 3. [note 3] Thus, for example, x 2 − 3y 2 = −1 is never solvable, but x 2 − 5y 2 = −1 may be. [27] The first few numbers n for which x 2 − ny 2 = −1 is solvable are
The result, x 2, is a "better" approximation to the system's solution than x 1 and x 0. If exact arithmetic were to be used in this example instead of limited-precision, then the exact solution would theoretically have been reached after n = 2 iterations ( n being the order of the system).