Question

i need help! i need to know how to solve this.
sqrt(7^5u^12v^20w^65x^80

Answer 1

\[\sqrt{7^{5}u ^{12} v ^{20} w ^{65} z ^{80}}\]

Answer 2

divide each exponent by the index, which is 2
so for example 2 goes in to 5 twice, with a remainder of 1
therefore \[\sqrt{7^5}=7^2\sqrt{7}\]

Answer 3

2 goes in to 12 six times with no remainder so
\[\sqrt{u^{12}}=u^6\]

Answer 4

similarly 2 goes in to 20 ten times with no remainder, so
\[\sqrt{v^{20}}=v^{10}\]

Answer 5

and so on

Answer 6

so everythin has to be divided into 2?

Answer 7

yes divide all the exponents by 2
if there is a remainder of 1, i.e. if the exponent is odd, one stays in and the rest comes out of radical

Answer 8

ok so \[\sqrt{w ^{65}}= w \sqrt{w ^{32}}\]

Answer 9

no not quite

Answer 10

how many times does two go in to 65?

Answer 11

oh i see the mistake
you have it backwards
\[\sqrt{w^{65}}=w^{32}\sqrt{w}\]

Answer 12

ohhhh ok. so i have to flip it. but in a way i was correct?

Answer 13

\(w^{32}\) comes out of the radical, one \(w\) stays in

Answer 14

ok!

Answer 15

x^40\[\sqrt{x}\]

Answer 16

\[x^{40}\sqrt{x}\]

Answer 17

how many times does 2 go in to 40?

Answer 18

20

Answer 19

is there a remainder?

Answer 20

im dividing it 40 by 2

Answer 21

yes divide 40 by 2
you get 20
but no remainder right?

Answer 22

no

Answer 23

this means
\[\sqrt{x^{40}}=x^{20}\]

Answer 24

since there is no remainder, nothing is left inside the radical
it all comes out

Answer 25

\[^{20}\sqrt{x}\]

Answer 26

satellite, your awesome. :)

Answer 27

x

Answer 28

(blush)

Answer 29

@mperez1312 if there is no remainder, nothing stays inside the radical

Answer 30

for example \(\sqrt{2^6}=2^3\) nothing is left in the radical
we see that \(\sqrt{2^6}=\sqrt{64}=8\) and \(2^3=8\) there is no radical because 2 goes in to 6 evenly

Answer 31

ok
perfect! i got it!

Answer 32

thank you for your help!