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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.
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 over F 16, where it has the two roots ab and ab + a, where b is a root of x 2 + x + a in F 16. This is a special case of Artin–Schreier theory.
Graph of a sextic function, with 6 real roots (crossings of the x axis) and 5 critical points. Depending on the number and vertical locations of minima and maxima, the sextic could have 6, 4, 2, or no real roots. The number of complex roots equals 6 minus the number of real roots. In algebra, a sextic (or hexic) polynomial is a polynomial of ...
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 ...
Solving an equation f(x) = g(x) is the same as finding the roots of the function h(x) = f(x) – g(x). Thus root-finding algorithms can be used to solve any equation of continuous functions. However, most root-finding algorithms do not guarantee that they will find all roots of a function, and if such an algorithm does not find any root, that ...
An example of a more complicated (although small enough to be written here) solution is the unique real root of x 5 − 5x + 12 = 0. Let a = √ 2φ −1, b = √ 2φ, and c = 4 √ 5, where φ = 1+ √ 5 / 2 is the golden ratio. Then the only real solution x = −1.84208... is given by
[4] This will not work if squares or higher power of x occurs in an exponent, or if the "base constants" do not "share" a common q. sometimes, substituting y=xe x may obtain an algebraic equation; after the solutions for y are known, those for x can be obtained by applying the Lambert W function, [citation needed] e.g.: