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One particular solution is x = 0, y = 0, z = 0. 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 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.
def f (x): return x ** 2-2 # f(x) = x^2 - 2 def f_prime (x): return 2 * x # f'(x) = 2x def newtons_method (x0, f, f_prime, tolerance, epsilon, max_iterations): """Newton's method Args: x0: The initial guess f: The function whose root we are trying to find f_prime: The derivative of the function tolerance: Stop when iterations change by less ...
If an equation P(x) = 0 of degree n has a rational root α, the associated polynomial can be factored to give the form P(X) = (X – α)Q(X) (by dividing P(X) by X – α or by writing P(X) – P(α) as a linear combination of terms of the form X k – α k, and factoring out X – α. Solving P(x) = 0 thus reduces to solving the degree n – 1 ...
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
You will find choosing a strategy increasingly easy. A partial list of strategies is included: Guess and check [9] Make an orderly list [10] Eliminate possibilities [11] Use symmetry [12] Consider special cases [13] Use direct reasoning; Solve an equation [14] Also suggested: Look for a pattern [15] Draw a picture [16] Solve a simpler problem ...
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For example, let a denote a multiplicative generator of the group of units of F 4, the Galois field of order four (thus a and a + 1 are roots of x 2 + x + 1 over F 4. Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits ...