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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.}
The nested radicals in this solution cannot in general be simplified unless the cubic equation has at least one rational solution. Indeed, if the cubic has three irrational but real solutions, we have the casus irreducibilis, in which all three real solutions are written in terms of cube roots of complex numbers. On the other hand, consider the ...
So, if one starts from a solution in terms of radicals, one gets an increasing sequence of fields such that the last one contains the solution, and each is a normal extension of the preceding one with a Galois group that is cyclic. Conversely, if one has such a sequence of fields, the equation is solvable in terms of radicals.
In general, there are four such points, giving four different solutions for the LLL Apollonius problem. The radius of each solution is determined by finding a point of tangency T, which may be done by choosing one of the three intersection points P between the given lines; and drawing a circle centered on the midpoint of C and P of diameter ...
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 equation, consisting generally of two expressions related by an equals sign.
Searching for solutions that belong to a specific set is a problem which is generally much more difficult, and is outside the scope of this article, except for the case of the solutions in a given finite field. For the case of solutions of which all components are integers or rational numbers, see Diophantine equation.
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Algebraic operations in the solution to the quadratic equation.The radical sign √, denoting a square root, is equivalent to exponentiation to the power of 1 / 2 .The ± sign means the equation can be written with either a + or a – sign.
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