Question

reverse the process. Un-simplify the simplified radical form. Write as one number under the radicand

3sqrt2
-5Sqrt6
plus or minus24sqrt21

Answer 1

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

Answer 2

how did u know to do that

Answer 3

what about the other ones do i do the same thing

Answer 4

@phi

Answer 5

\[ -5 \sqrt{6} \]
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:
\[ -\sqrt{150} \]

Answer 6

can you do the last one?

Answer 7

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

Answer 8

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

Answer 9

yes

Answer 10

prime factorization

Answer 11

right :D

Answer 12

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

Answer 13

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

Answer 14

is this correct \[ab\]

Answer 15

absqrta

Answer 16

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)

Answer 17

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.

\[ \sqrt{a^{20}} = a^{10} \]

Answer 18

ahhh ok gocha

Answer 19

thank u

Answer 20

yw