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For solving the cubic equation x 3 + m 2 x = n where n > 0, Omar Khayyám constructed the parabola y = x 2 /m, the circle that has as a diameter the line segment [0, n/m 2] on the positive x-axis, and a vertical line through the point where the circle and the parabola intersect above the x-axis.
Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. There is a unique plane in three-dimensional space which passes through the three points with these coordinates, and this plane is the set of all points whose coordinates are solutions of the equation.
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]
Vertical line of equation x = a Horizontal line of equation y = b. Each solution (x, y) of a linear equation + + = may be viewed as the Cartesian coordinates of a point in the Euclidean plane. With this interpretation, all solutions of the equation form a line, provided that a and b are not both zero. Conversely, every line is the set of all ...
If we solve this equation, we find that x = 2. More generally, we find that + + + + is the positive real root of the equation x 3 − x − n = 0 for all n > 0. For n = 1, this root is the plastic ratio ρ, approximately equal to 1.3247.
Microsoft Math 1.0: Part of Microsoft Student 2006 Microsoft Math 2.0 : Part of Microsoft Student 2007 Microsoft Math 3.0 : Standalone commercial product that requires product activation ; includes calculus support, digital ink recognition features and a special display mode for video projectors
The roots, stationary points, inflection point and concavity of a cubic polynomial x 3 − 6x 2 + 9x − 4 (solid black curve) and its first (dashed red) and second (dotted orange) derivatives. The critical points of a cubic function are its stationary points, that is the points where the slope of the function is zero. [2]
Thomas' algorithm is not stable in general, but is so in several special cases, such as when the matrix is diagonally dominant (either by rows or columns) or symmetric positive definite; [1] [2] for a more precise characterization of stability of Thomas' algorithm, see Higham Theorem 9.12. [3]