College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Chapter P Review - Exercises - Page 77: 98


True for $a\ne1$.

Work Step by Step

This is mostly true. We can show that the left equation turns into the right equation by multiplying the numerator and denominator by $1-\sqrt{a}$ and using the fact that $(a-b)(a+b)=a^2-b^2$: $\displaystyle\frac{1+\sqrt{a}}{1-a}=\frac{1+\sqrt{a}}{1-a}\cdot\frac{1-\sqrt{a}}{1-\sqrt{a}}=\frac{(1-a)}{(1-a)(1-\sqrt{a})}=\frac{1}{1-\sqrt{a}}$ However, $a$ can not be $1$ because then the denominator would be $0$ and the equation would be undefined. Thus the equation is true for $a\ne1$.
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