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Mathematics 19 Online
OpenStudy (anonymous):

How would I simplify this radical? I have the answer but I need to know the steps taken.

OpenStudy (anonymous):

\[3\sqrt{12} * \sqrt{6}\]

jimthompson5910 (jim_thompson5910):

hint: \[\Large \sqrt{x}*\sqrt{y} = \sqrt{x*y}\]

OpenStudy (anonymous):

\[3\sqrt{12 * 6} = 3\sqrt{72}\] Like this?

jimthompson5910 (jim_thompson5910):

yes

jimthompson5910 (jim_thompson5910):

now you can factor 72 in such a way where one of the factors is a perfect square

jimthompson5910 (jim_thompson5910):

and then you would use that rule in reverse

OpenStudy (anonymous):

I know that 9 times 8 is 72. Is that important?

jimthompson5910 (jim_thompson5910):

you can use a larger perfect square actually

jimthompson5910 (jim_thompson5910):

but you are on the right track

jimthompson5910 (jim_thompson5910):

factors of 72: 1,2,3,4,6,8,9,12,18,24,36,72

OpenStudy (phi):

I would factor the numbers before multiplying together. (It's easier) \[ 3\sqrt{12} \cdot \sqrt{6} = 3 \cdot \sqrt{3\cdot 2\cdot 2}\cdot \sqrt{2\cdot 3}\] you can combine the 2 square roots, as jim posted up above \[ 3 \cdot \sqrt{3\cdot 2\cdot 2\cdot 2\cdot 3}\] now look for *pairs" of numbers inside the square root, and "pull them out" there is is pair of 3's and one pair of 2's (leaving 1 2 left over that stays inside) when we pull out a pair, it becomes just a single number on the outside we get \[ 3 \cdot 3 \cdot 2 \sqrt{2} \] or \[ 18 \sqrt{2} \]

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