How do you combine the radical $\sqrt{6} - \sqrt{27} + 2 \sqrt{54} + 3 \sqrt{48}$ $\sqrt{6}-\sqrt{27}+2\sqrt{54}+3\sqrt{48}$ ?

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Answer 1 · 2014-11-05T18:30:01

$\sqrt{2} - \sqrt{27} + 2 \sqrt{54} + 3 \sqrt{48}$ $sqrt{2}-sqrt{27}+2sqrt{54}+3sqrt{48}$

by $27 = 3^{3}$ $27=3^3$ , $54 = 2 \cdot 3^{3}$ $54=2 cdot 3^3$ , and $48 = 2^{4} \cdot 3$ $48=2^4 cdot 3$ ,

$= \sqrt{6} - \sqrt{3^{3}} + 2 \sqrt{2 \cdot 3^{3}} + 3 \sqrt{2^{4} \cdot 3}$ $=sqrt{6}-sqrt{3^3}+2sqrt{2 cdot 3^3}+3sqrt{2^4 cdot 3}$

by pulling out factors out of the square-roots,

$= \sqrt{6} - 3 \sqrt{3} + 6 \sqrt{6} + 12 \sqrt{3}$ $=sqrt{6}-3sqrt{3}+6sqrt{6}+12sqrt{3}$

by combining the like terms,

$= 7 \sqrt{6} + 9 \sqrt{3}$ $=7sqrt{6}+9sqrt{3}$

I hope that this was helpful.

How do you combine the radical \sqrt{6}-\sqrt{27}+2\sqrt{54}+3\sqrt{48}√6−√27+2√54+3√48\sqrt{6}-\sqrt{27}+2\sqrt{54}+3\sqrt{48}?