Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-3 Operations with Radial Expressions - Lesson Check: 8

Answer

The mistake is that $\sqrt 3 \times \sqrt 3$ is $\sqrt 9$. The square root of 9 is 3 not 9 because 3 x 3 = 9 The correct answer is $\frac{\sqrt 3 - 1}{2}$

Work Step by Step

$\frac{1}{\sqrt 3 -1}$ We multiply the faraction by the conjugate of $\sqrt 3 -1$ which is $\sqrt 3 +1$ $\frac{1}{\sqrt 3 -1} \times \frac{\sqrt 3 +1}{\sqrt 3 +1}$ We multiply the numerator by the numerator and the denominator by the denominator. $\frac{1 \times \sqrt 3 +1}{\sqrt 3 -1 \times \sqrt 3 +1}$ $\frac{\sqrt 3 - 1}{\sqrt 9 - 1}$ Square root of 9 is 3 because 3 x 3 = 9 $\frac{\sqrt 3 - 1}{3 - 1}$ $\frac{\sqrt 3 - 1}{2}$ The mistake is that $\sqrt 3 \times \sqrt 3$ is $\sqrt 9$. The square root of 9 is 3 not 9 because 3 x 3 = 9
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