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The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras. [1]
Note that even simple equations like = are solved using cross-multiplication, since the missing b term is implicitly equal to 1: =. Any equation containing fractions or rational expressions can be simplified by multiplying both sides by the least common denominator.
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
But if is substituted for in the original equation, the result is the invalid equation =. This counterintuitive result occurs because in the case where x = 0 {\displaystyle x=0} , multiplying both sides by x {\displaystyle x} multiplies both sides by zero, and so necessarily produces a true equation just as in the first example.
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.
Staffing or maintenance problems are considered controllable, for instance. But a hurricane obviously isn’t. So in that case the airline would not have the same obligations to the traveler.
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