Ad
related to: how to check for extraneous solutions in radicals practice test questions
Search results
Results from the WOW.Com Content Network
Therefore, the solution = is extraneous and not valid, and the original equation has no solution. For this specific example, it could be recognized that (for the value x = − 2 {\displaystyle x=-2} ), the operation of multiplying by ( x − 2 ) ( x + 2 ) {\displaystyle (x-2)(x+2)} would be a multiplication by zero.
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
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 ...
Plot of the Bring radical for real argument. In algebra, the Bring radical or ultraradical of a real number a is the unique real root of the polynomial + +.. The Bring radical of a complex number a is either any of the five roots of the above polynomial (it is thus multi-valued), or a specific root, which is usually chosen such that the Bring radical is real-valued for real a and is an ...
Casus irreducibilis (from Latin 'the irreducible case') is the name given by mathematicians of the 16th century to cubic equations that cannot be solved in terms of real radicals, that is to those equations such that the computation of the solutions cannot be reduced to the computation of square and cube roots.
Radical extensions occur naturally when solving polynomial equations in radicals. In fact a solution in radicals is the expression of the solution as an element of a radical series: a polynomial f over a field K is said to be solvable by radicals if there is a splitting field of f over K contained in a radical extension of K.
The solutions of the system are in one-to-one correspondence with the roots of h and the multiplicity of each root of h equals the multiplicity of the corresponding solution. The solutions of the system are obtained by substituting the roots of h in the other equations. If h does not have any multiple root then g 0 is the derivative of h.
Then any solution of the inhomogeneous equation may have a solution of the homogeneous equation added to it, and still remain a solution. For example in mathematical physics , the homogeneous equation may correspond to a physical theory formulated in empty space , while the inhomogeneous equation asks for more 'realistic' solutions with some ...
Ad
related to: how to check for extraneous solutions in radicals practice test questions