## College Algebra (11th Edition)

$r-2+\dfrac{1}{r}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(r^{1/2}-r^{-1/2})^2 ,$ use the special product on squaring binomials. Then, use the laws of exponents to simplify the resulting expression. $\bf{\text{Solution Details:}}$ Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (r^{1/2})^2-2(r^{1/2})(r^{-1/2})+(r^{-1/2})^2 .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} r^{\frac{1}{2}\cdot2}-2(r^{1/2})(r^{-1/2})+r^{-\frac{1}{2}\cdot2} \\\\= r^{1}-2(r^{1/2})(r^{-1/2})+r^{-1} \\\\= r-2(r^{1/2})(r^{-1/2})+r^{-1} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} r-2r^{\frac{1}{2}+\left(-\frac{1}{2}\right)}+r^{-1} \\\\= r-2r^{\frac{1}{2}-\frac{1}{2}}+r^{-1} \\\\= r-2r^{0}+r^{-1} .\end{array} Since any expression raised to the power of zero is $1$, the expression above is equivalent to \begin{array}{l}\require{cancel} r-2(1)+r^{-1} \\\\= r-2+r^{-1} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} r-2+\dfrac{1}{r^1} \\\\= r-2+\dfrac{1}{r} .\end{array}