Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 9 - Solving Quadratic Equations - 9.1 - Properties of Radicals - Exercises - Page 486: 44

Answer

$\frac{\sqrt{3}-\sqrt {7}}{\sqrt {3}-\sqrt {7}}$.

Work Step by Step

The conjugate of $\sqrt {3}+\sqrt {7}$ is $\sqrt {3}-\sqrt {7}$. Appropriate form of $1$ that can be used as a factor to eliminate the radical in the denominator is $\frac{\sqrt{3}-\sqrt {7}}{\sqrt {3}-\sqrt {7}}$. The denominator becomes: $(\sqrt 3+\sqrt 7)(\sqrt 3-\sqrt 7)=(\sqrt 3)^2-(\sqrt 7)^2=3-7=-4$
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