Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises - Page 34: 130

Answer

$\dfrac{1}{2}x^{-1/2}(3x+4)^{1/2}-\dfrac{3}{2}x^{1/2}(3x+4)^{-1/2}=\dfrac{2}{\sqrt{3x^{2}+4x}}$

Work Step by Step

$\dfrac{1}{2}x^{-1/2}(3x+4)^{1/2}-\dfrac{3}{2}x^{1/2}(3x+4)^{-1/2}$ Take out common factor $\dfrac{1}{2}x^{-1/2}(3x+4)^{-1/2}$: $\dfrac{1}{2}x^{-1/2}(3x+4)^{1/2}-\dfrac{3}{2}x^{1/2}(3x+4)^{-1/2}=...$ $...=\dfrac{1}{2}x^{-1/2}(3x+4)^{-1/2}\Big[(3x+4)-3x\Big]=...$ Simplify the expression inside brackets: $...=\dfrac{1}{2}x^{-1/2}(3x+4)^{-1/2}(4)=2x^{-1/2}(3x+4)^{-1/2}=...$ Rearrange: $...=\dfrac{2}{x^{1/2}(3x+4)^{1/2}}=\dfrac{2}{(\sqrt{x})(\sqrt{3x+4})}=\dfrac{2}{\sqrt{3x^{2}+4x}}$
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