College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises - Page 68: 96

Answer

$\dfrac{3\sqrt{5} + 3\sqrt{15} - 2\sqrt 3 - 6}{33}$

Work Step by Step

First it can be easily spotted that the denominator radicals could be simplified by the difference of two squares, therefore the numerator and denominator will be multiplied by $\left(3\sqrt 5 - 2\sqrt 3\right)$ to obtain: $$=\dfrac{(3\sqrt 5 - 2\sqrt 3).(1 + \sqrt 3)}{(3\sqrt 5 - 2\sqrt 3).(3\sqrt 5 + 2\sqrt 3)}$$ By expanding the brackets the denominaotr becomes: \begin{align*}&=\dfrac{(3\sqrt 5 - 2\sqrt 3)(1 + \sqrt 3)}{(3\sqrt 5)^{2} - (2\sqrt 3)^2}\\ \\&=\dfrac{(3\sqrt 5 - 2\sqrt 3)(1 + \sqrt 3)}{9\cdot5-4\cdot3}\\ \\&=\dfrac{(3\sqrt 5 - 2\sqrt 3)(1 + \sqrt 3)}{9\cdot5-4\cdot3} \\&=\dfrac{(3\sqrt 5 - 2\sqrt 3)(1 + \sqrt 3)}{45-12} \\&=\dfrac{(3\sqrt 5 - 2\sqrt 3)(1 + \sqrt 3)}{33} \end{align*} The denominator now is free of radicals and is in the simplest form, the next step is to expand and simplify the numerator. \begin{align*} &=\frac{3\sqrt 5(1+\sqrt3)-2\sqrt3(1+\sqrt3)}{(9.5) - (4.3)}\\ \\&=\frac{3\sqrt 5 + 3\sqrt{15} - 2\sqrt 3 - 2\sqrt 9}{(9.5) - (4.3)}\\ \\&=\frac{3\sqrt 5 + 3\sqrt{15} - 2\sqrt 3 - 2(3)}{33}\\ \\&=\frac{3\sqrt 5 + 3\sqrt{15} - 2\sqrt 3 - 6}{33} \end{align*} The final answer then becomes: $$=\frac{3\sqrt 5 + 3\sqrt{15} - 2\sqrt 3 - 6}{33}$$
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.