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 - Practice and Problem-Solving Exercises - Page 616: 34



Work Step by Step

We must multiply by the conjugate (change the sign between the two terms ) of the denominator. $\frac{-1}{2-2\sqrt{3}}\cdot \frac{2+2\sqrt{3}}{2+2\sqrt{3}}=\frac{-2-2\sqrt{3}}{4+4\sqrt{3}-4\sqrt{3}-(2\sqrt{3})^2}=\frac{-2-2\sqrt{3}}{4-4(3)}=\frac{-2-2\sqrt{3}}{-8}=\frac{1+\sqrt{3}}{4}$
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