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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]
The Algebra is perhaps Abu Kamil's most influential work, which he intended to supersede and expand upon that of Al-Khwarizmi. [2] [11] Whereas the Algebra of al-Khwarizmi was geared towards the general public, Abu Kamil was addressing other mathematicians, or readers familiar with Euclid's Elements. [11]
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 ...
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.
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 ...
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).
The question is whether or not, for all problems for which an algorithm can verify a given solution quickly (that is, in polynomial time), an algorithm can also find that solution quickly. Since the former describes the class of problems termed NP, while the latter describes P, the question is equivalent to asking whether all problems in NP are ...
The solution = is in fact a valid solution to the original equation; but the other solution, =, has disappeared. The problem is that we divided both sides by x {\displaystyle x} , which involves the indeterminate operation of dividing by zero when x = 0. {\displaystyle x=0.}