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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.
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.
If a quadratic function is equated with zero, then the result is a quadratic equation. The solutions of a quadratic equation are the zeros (or roots) of the corresponding quadratic function, of which there can be two, one, or zero. The solutions are described by the quadratic formula. A quadratic polynomial or quadratic function can involve ...
All quadratic equations have exactly two solutions in complex numbers (but they may be equal to each other), a category that includes real numbers, imaginary numbers, and sums of real and imaginary numbers. Complex numbers first arise in the teaching of quadratic equations and the quadratic formula. For example, the quadratic equation
(These can be quickly shown to be true by direct substitution into the quartic above and noting that ω 6 = ω, and ω 7 = ω 2.) So p 1 and p 2 are the roots of the quadratic equation x 2 + x − 1 = 0. The Carlyle circle associated with this quadratic has a diameter with endpoints at (0, 1) and (−1, −1) and center at (−1/2, 0).
By the fundamental theorem of algebra, if the monic polynomial equation x 2 + bx + c = 0 has complex coefficients, it must have two (not necessarily distinct) complex roots. Unfortunately, the discriminant b 2 − 4c is not as useful in this situation, because it may be a complex number. Still, a modified version of the general theorem can be ...
One way to see this is to note that the graph of the function f(x) = x 2 is a parabola whose vertex is at the origin (0, 0). Therefore, the graph of the function f ( x − h ) = ( x − h ) 2 is a parabola shifted to the right by h whose vertex is at ( h , 0), as shown in the top figure.
The quadratic formula, the symbolic solution of the quadratic equation ax 2 + bx + c = 0 An example of using Newton–Raphson method to solve numerically the equation f ( x ) = 0 In mathematics , to solve an equation is to find its solutions , which are the values ( numbers , functions , sets , etc.) that fulfill the condition stated by the ...