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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.
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.
When the monic quadratic equation with real coefficients is of the form x 2 = c, the general solution described above is useless because division by zero is not well defined. As long as c is positive, though, it is always possible to transform the equation by subtracting a perfect square from both sides and proceeding along the lines ...
That is, h is the x-coordinate of the axis of symmetry (i.e. the axis of symmetry has equation x = h), and k is the minimum value (or maximum value, if a < 0) of the quadratic function. 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).
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]
A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.). A well-known example is the quadratic formula
Solving an equation symbolically means that expressions can be used for representing the solutions. 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 ...
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