Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - Chapter Test - Page 427: 6

Answer

$20\sqrt {3}$

Work Step by Step

Before we can add these two numbers, we must transform them so that the radicands are the same. Only then can we perform the operation we are asked to do. Let's expand the radicands in all of the radicals so that they are rewritten as the product of a perfect square and another number: First term: $\sqrt {48}$ = $\sqrt {16\cdot3}$ Second term: $2\sqrt {27}$ = $2\sqrt {9\cdot3}$ Third term: $5\sqrt {12}$ = $5\sqrt {4\cdot3}$ Let's put them all back into the expression: $\sqrt {16\cdot3} + 2\sqrt {9\cdot3} + 5\sqrt {4\cdot3}$ Let's rewrite all the perfect squares as a base raised to the second power so that we can take it out from under the radical sign later: $\sqrt {4^2\cdot3} + 2\sqrt {3^2\cdot3} + 5\sqrt {2^2\cdot3}$ We can take the square roots of all the perfect squares: $4\sqrt {3} + 2\cdot3\sqrt {3} + 5\cdot2\sqrt {3}$ Multiply the coefficients: $4\sqrt {3} + 6\sqrt {3} + 10\sqrt {3}$ Finally, we can perform the operation in the original problem because we now have the same radicand and indices for all terms. We just add the coefficients from left to right, keeping the radical as-is: $10\sqrt {3} + 10\sqrt {3}$ Add one more time: $20\sqrt {3}$
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