Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 96

Answer

$\sqrt{x+1}-\sqrt{x}=\dfrac{1}{\sqrt{x+1}+\sqrt{x}}$

Work Step by Step

$\sqrt{x+1}-\sqrt{x}$ Multiply this expression by $\dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}$ and simplify: $\sqrt{x+1}-\sqrt{x}=(\sqrt{x+1}-\sqrt{x})\Big(\dfrac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\Big)=...$ $...=\dfrac{(\sqrt{x+1})^{2}-(\sqrt{x})^{2}}{\sqrt{x+1}+\sqrt{x}}=\dfrac{x+1-x}{\sqrt{x+1}+\sqrt{x}}=\dfrac{1}{\sqrt{x+1}+\sqrt{x}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.