Chapter A.10, Problem 95AYU

To determine

Express the expression $\frac{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x{\left(x+4\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{x+4}$ as a single quotient in which only positive exponent or radical appear.

Answer to Problem 95AYU

The required form of the given expression $\frac{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x{\left(x+4\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{x+4}$ is $\frac{4-x}{{\left(x+4\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$.

Explanation of Solution

Given:

It is asked to determine the expression as a single quotient in which only positive exponents or radicals appear.

Calculation:

Consider $\frac{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x{\left(x+4\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{x+4}$,

$\frac{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x{\left(x+4\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{x+4}=\frac{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{2x}{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{x+4}$

Simplify the numerator expression.

Now consider ${\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{2x}{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$,

${\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{2x}{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{\left(x+4\right)-2x}{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}= \frac{4-x}{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

$\frac{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2x{\left(x+4\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{x+4}=\frac{\frac{4-x}{{\left(x+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{x+4}=\frac{4-x}{{\left(x+4\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$