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Mathematics 48 Online
OpenStudy (anonymous):

reverse the process. Un-simplify the simplified radical form. Write as one number under the radicand 3sqrt2 -5Sqrt6 plus or minus24sqrt21

OpenStudy (anonymous):

i guess you could write \(3\sqrt{2}=\sqrt{3^2\times 2}=\sqrt{9\times 2}=\sqrt{18}\)

OpenStudy (anonymous):

how did u know to do that

OpenStudy (anonymous):

what about the other ones do i do the same thing

OpenStudy (anonymous):

@phi

OpenStudy (phi):

\[ -5 \sqrt{6} \] they want you to move the 5 (not the minus sign!) inside the square root so first square the 5: write 5*5 and move this pair inside the square root \[ -\sqrt{5*5*6} \] notice that to undo this, we would find the pair 5*5, take them out of the square root and get \(-5\sqrt{6} \) . That is what you did in the earlier questions. But back to this one. multiply the numbers inside to get the final answer: \[ -\sqrt{150} \]

OpenStudy (phi):

can you do the last one?

OpenStudy (anonymous):

it \[\pm \sqrt{12096}\]

OpenStudy (phi):

yes. Could you undo this monster? (get back to ±24 sqrt(21)) ??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

prime factorization

OpenStudy (anonymous):

right :D

OpenStudy (phi):

Good. If you do that one, you are doing great.

OpenStudy (anonymous):

what about this one \[\sqrt{a ^{3}b ^{2}}\]

OpenStudy (anonymous):

is this correct \[ab\]

OpenStudy (anonymous):

absqrta

OpenStudy (phi):

same idea as numbers. b^2 means b*b, a^3 mean a*a*a so look for pairs: b*b*a*a *a take out the pairs to get a*b sqrt(a)

OpenStudy (phi):

of course, with exponents the fast way is to look for even exponents b^2 becomes b after you take its square root. if you have odd exponents like a^3 write it as a^2*a and then take out the a^2. \[ \sqrt{a^{20}} = a^{10} \]

OpenStudy (anonymous):

ahhh ok gocha

OpenStudy (anonymous):

thank u

OpenStudy (phi):

yw

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