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Algebra 19 Online
OpenStudy (anonymous):

How to solve: sqrt3 x ( sqrt2 +sqrt6 ) answerbook says: sqrt6 + 3 sqrt2

OpenStudy (lxelle):

=Sqrt 3 x sqrt(12) =Sqrt 24 =sqrt 6 + sqrt 18 =sqrt 6 + sqrt (9x2) =sqrt 6 + 3sqrt2

OpenStudy (phi):

\[ \sqrt{3} ( \sqrt{2} + \sqrt{6} )\] distribute the sqr(3): \[ \sqrt{3} \sqrt{2} + \sqrt{3} \sqrt{6} \] use this "rule" \[ \sqrt{a\cdot b} = \sqrt{a} \sqrt{b} \] and vice versa \[ \sqrt{3} ( \sqrt{2} + \sqrt{6} )\] to write the first term as sqrt(6) and the second term as sqrt(18): \[ \sqrt{6}+\sqrt{18} \] notice 18 = 9*2 so we we can write sqr(18) as sqr(9) * sqr(2) (using the rule "the other way" \[ \sqrt{6}+\sqrt{9}\sqrt{2} \] sqr(9)= sqr(3*3) = 3, so \[ \sqrt{6}+3\sqrt{2} \]

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