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In mathematics, a quadratic equation (from Latin quadratus 'square') is an equation that can be rearranged in standard form as [1] + + =, where the variable x represents an unknown number, and a, b, and c represent known numbers, where a ≠ 0. (If a = 0 and b ≠ 0 then the equation is linear, not quadratic
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]
Given a quadratic polynomial of the form + + it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial. Example: + + = [+ +] = [(+) +] = (+) + = (+) + This process of factoring out the coefficient a can further be simplified by only factorising it out of the first 2 terms.
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 ...
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 ...
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). Carlyle circles are used to construct p 1 and p 2. From the definitions of p 1 and p 2 it also follows that p 1 = 2 cos(2 π /5), p 2 = 2 ...
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
It follows from the formula that r is the quotient of two polynomials of degree s if the method has s stages. Explicit methods have a strictly lower triangular matrix A, which implies that det(I − zA) = 1 and that the stability function is a polynomial. [32] The numerical solution to the linear test equation decays to zero if | r(z) | < 1 ...
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