Question

whats the square root of 16x^4y^7

Answer 1

this can be rewritten as

\[\sqrt{16} \times \sqrt{x^4} \times \sqrt{y^6}\times \sqrt{y}\]

can you simplify it from here

Answer 2

no

Answer 3

do you know any of the index laws...?

Answer 4

no

Answer 5

makes it hard to do this question

Answer 6

can you solve it and then show me

Answer 7

hello

Answer 8

http://www.algebra.com/algebra/homework/Square-cubic-other-roots/Square-cubic-other-roots.faq.question.325222.html here's an example

Answer 9

can you simplify

\[\sqrt{16} = \sqrt{4^2} =?\]

Answer 10

so is it 4x squared y to the fourth

Answer 11

helloo

Answer 12

nope.... you just need to get the powers of x and y correct its \[4x^?y^? \sqrt{y}\]

Answer 13

thats so confusing i still dont get it

Answer 14

use this to see if you can get the power of x

\[\sqrt{y^6} = y^3\]

Answer 15

x^2 and y^4

Answer 16

nope... \[\sqrt{16} timessqrt{x^4} \times \sqrt{y^3} \times \sqrt{y} = 4x^2y^3\sqrt{y}\]

Answer 17

thats makes no sense

Answer 18

I think the problem is that you don't seem to have learnt about power laws or radical laws

Answer 19

so if i have a problem like \[\sqrt{60a ^{6}b ^{5}c ^{3}}\]

Answer 20

its 2a^3 b^4 c2 root 15bc

Answer 21

not quite a is correct.... the

b needs to be split into an even and odd power

\[b^5 = b^4 \times b\]

same for c

\[c^3 = c^2 \times c\]

and split 60 into the product of a square number and another number

\[60 = 4\times15.... = 2^2 \times 15\]

so the problem can be split into parts

\[\sqrt{2^2 a^6b^4c^2} \times \sqrt{15bc}\]

the 1st square root just write the the same number and letters but just halve the powers

the 2nd square root won't change

Answer 22

so it would be 4a to the 3 b to the 4 and c 2 root 15bc

Answer 23

bits are correct

\[\sqrt{2^2a^6b^4c^2} = 2a^3b^2c \]

Answer 24

so its 2a3b2c root 15bc

Answer 25

thats right

Answer 26

omg thanks i understand it now

Answer 27

hope it helps

Answer 28

its does alot

Answer 29

i know you guys already got there but i did this for extra :P