reverse the process. Un-simplify the simplified radical form. Write as one number under the radicand 3sqrt2 -5Sqrt6 plus or minus24sqrt21
i guess you could write \(3\sqrt{2}=\sqrt{3^2\times 2}=\sqrt{9\times 2}=\sqrt{18}\)
how did u know to do that
what about the other ones do i do the same thing
@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: \[ -\sqrt{150} \]
can you do the last one?
it \[\pm \sqrt{12096}\]
yes. Could you undo this monster? (get back to ±24 sqrt(21)) ??
yes
prime factorization
right :D
Good. If you do that one, you are doing great.
what about this one \[\sqrt{a ^{3}b ^{2}}\]
is this correct \[ab\]
absqrta
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)
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. \[ \sqrt{a^{20}} = a^{10} \]
ahhh ok gocha
thank u
yw
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