College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 50: 117

Answer

$\frac{w}{h} = \frac{\sqrt 5 + 1}{2}$

Work Step by Step

$ \frac{w}{h}$ = $\frac{2}{\sqrt 5 - 1}$ To rationalize the denominator multiply the numerator and denominator by the conjugate of the denominator. The conjugate of the denominator $\sqrt 5 - 1 $ is $\sqrt 5 + 1 $ = $\frac{2}{\sqrt 5 - 1}$ $\times$ $\frac{\sqrt 5 + 1}{\sqrt 5 + 1}$ = $ \frac{2(\sqrt 5 + 1)}{(\sqrt 5 -1)(\sqrt 5 + 1)}$ $( \sqrt a - \sqrt b)( \sqrt a + \sqrt b)$ = $ (\sqrt a)^{2}$ - $ (\sqrt b)^{2}$. Therefore, $( \sqrt 5- 1)( \sqrt5 +1)$ = $ (\sqrt5)^{2}$ - $ (1)^{2}$. $= \frac{2(\sqrt 5 + 1)}{ (\sqrt5)^{2} - (1)^{2}}$ $= \frac{2(\sqrt 5 + 1)}{ (5 - 1)}$ $= \frac{2(\sqrt 5 + 1)}{4}$ $ = \frac{2}{4} \times (\sqrt 5 + 1)$ $\frac{w}{h}= \frac{\sqrt 5 + 1}{2}$ $\sqrt 5 \approx 2.2361$ $\frac{w}{h}= \frac{2.2361 + 1}{2}$ $\frac{w}{h}= \frac{3.2361}{2} = 1.61805$ $\frac{w}{h}= 1.62$ (to nearest hundredth) $\frac{w}{h}= \frac{1.62}{1}$ Ratio of width to height is 1.62 to 1
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