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The other roots of the equation are obtained either by changing of cube root or, equivalently, by multiplying the cube root by a primitive cube root of unity, that is . This formula for the roots is always correct except when p = q = 0 , with the proviso that if p = 0 , the square root is chosen so that C ≠ 0 .
The solutions of this equation are the x-values of the critical points and are given, using the quadratic formula, by =. The sign of the expression Δ 0 = b 2 – 3ac inside the square root determines the number of critical points. If it is positive, then there are two critical points, one is a local maximum, and the other is a local minimum.
Cubic equations, which are polynomial equations of the third degree (meaning the highest power of the unknown is 3) can always be solved for their three solutions in terms of cube roots and square roots (although simpler expressions only in terms of square roots exist for all three solutions, if at least one of them is a rational number).
The definition: A real number is algebraic if it’s the root of some polynomial with integer coefficients. For example, x²-6 is a polynomial with integer coefficients, since 1 and -6 are integers.
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 ...
If the rational root test finds no rational solutions, then the only way to express the solutions algebraically uses cube roots. But if the test finds a rational solution r, then factoring out (x – r) leaves a quadratic polynomial whose two roots, found with the quadratic formula, are the remaining two roots of the cubic, avoiding cube roots.
is irreducible, because if it could be factored there would be a linear factor giving a rational solution, while none of the possible roots given by the rational root test are actually roots. Since its discriminant is positive, it has three real roots, so it is an example of casus irreducibilis. These roots can be expressed as