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Radical expression … - QuestionCove
OpenStudy (anonymous):

Radical expression question

5 years ago
OpenStudy (anonymous):

5 years ago
OpenStudy (anonymous):

@jim_thompson5910 , help please ?

5 years ago
OpenStudy (jim_thompson5910):

\[\Large \sqrt{6}\left(7\sqrt{3}+6\right)\] Start off by distributing to get \[\Large \sqrt{6}*7\sqrt{3}+\sqrt{6}*6\] \[\Large 7\sqrt{6}*\sqrt{3}+6\sqrt{6}\] What's next?

5 years ago
OpenStudy (anonymous):

combining like terms ?

5 years ago
OpenStudy (jim_thompson5910):

close, but what do you get when you simplify \[\Large \sqrt{6}*\sqrt{3}\] ??

5 years ago
OpenStudy (anonymous):

√2

5 years ago
OpenStudy (jim_thompson5910):

\[\Large \sqrt{6}*\sqrt{3}\] \[\Large \sqrt{6*3}\] \[\Large \sqrt{18}\] so far so good?

5 years ago
OpenStudy (anonymous):

So it's (sqrt) 18 instead of (sqrt) 2 ?

5 years ago
OpenStudy (jim_thompson5910):

close, but we're not done yet

5 years ago
OpenStudy (jim_thompson5910):

\[\Large \sqrt{18}\] \[\Large \sqrt{9*2}\] \[\Large \sqrt{9}*\sqrt{2}\] \[\Large 3\sqrt{2}\] ------------------------------ So \[\Large \sqrt{6}*\sqrt{3} = 3\sqrt{2}\]

5 years ago
OpenStudy (jim_thompson5910):

\[\Large \sqrt{6}\left(7\sqrt{3}+6\right)\] \[\Large \sqrt{6}*7\sqrt{3}+\sqrt{6}*6\] \[\Large 7\sqrt{6}*\sqrt{3}+6\sqrt{6}\] \[\Large 7*3\sqrt{2}+6\sqrt{6}\] \[\Large 21\sqrt{2}+6\sqrt{6}\] =================================== So in the end, \[\Large \sqrt{6}\left(7\sqrt{3}+6\right) = 21\sqrt{2}+6\sqrt{6}\]

5 years ago
OpenStudy (anonymous):

I see.

5 years ago
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