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Figure 1. Plots of quadratic function y = ax 2 + bx + c, varying each coefficient separately while the other coefficients are fixed (at values a = 1, b = 0, c = 0). A quadratic equation whose coefficients are real numbers can have either zero, one, or two distinct real-valued solutions, also called roots.
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 mathematics, a quadratic function of a single variable is a function of the form [1] = + +,,where is its variable, and , , and are coefficients.The expression + + , especially when treated as an object in itself rather than as a function, is a quadratic polynomial, a polynomial of degree two.
A finite-dimensional vector space with a quadratic form is called a quadratic space. The map Q is a homogeneous function of degree 2, which means that it has the property that, for all a in K and v in V : Q ( a v ) = a 2 Q ( v ) . {\displaystyle Q(av)=a^{2}Q(v).}
The quadratic programming problem with n variables and m constraints can be formulated as follows. [2] Given: a real-valued, n-dimensional vector c, an n×n-dimensional real symmetric matrix Q, an m×n-dimensional real matrix A, and; an m-dimensional real vector b, the objective of quadratic programming is to find an n-dimensional vector x ...
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The four roots of the depressed quartic x 4 + px 2 + qx + r = 0 may also be expressed as the x coordinates of the intersections of the two quadratic equations y 2 + py + qx + r = 0 and y − x 2 = 0 i.e., using the substitution y = x 2 that two quadratics intersect in four points is an instance of Bézout's theorem.