Question

Can someone please walk me through how to do square roots like breaking it down into a a number and a radical? Can you help me by giving me a few sample questions and walking me through it please? thanks!

Answer 1

do you mean \[\huge \sqrt{48} \implies \sqrt{16 \times 3} \implies 4\sqrt 3\]
something like that? or no?

Answer 2

yes something like that! please help

Answer 3

i think one of these two people typing something will give something helpful..

Answer 4

factorise the number in to a product of primes

\[48=2\times 24=2^2\times12=2^3\times6=2^4\times3=(2^2)^2\times3\]
\[\sqrt{48}=\sqrt{(2^2)^2\times3}=2^2\sqrt3=4\sqrt3\]

Answer 5

@sparky16 try checking this link out then tell me if it helps or not openstudy.com/updates/4fbae0ebe4b05565342fe421

Answer 6

http://openstudy.com/updates/4fbae0ebe4b05565342fe421

Answer 7

^there

Answer 8

So if what @lgbasallote and @UnkleRhaukus wrote isn't exactly what you were getting at the missing step is:\[\large\sqrt{48}\Rightarrow \sqrt{16*3}\Rightarrow\sqrt{4^2*3}\Rightarrow\sqrt{4^2}*\sqrt{3}\Rightarrow 4*\sqrt{3}\]

Answer 9

hehe i was just asking about the question...i wasnt going in any details but good eye in seeing the missing step :D

Answer 10

ok to clarify.. u pull out a group of two numbers and if there are four number you times them together to put on the outside? and what if there are 3 groups of two numbers... i dont know if this is possible but like 2 groups of 2's and one group of 3's

Answer 11

lets see if you can do this one\[\sqrt{8100}=\]

Answer 12

step 1 would be to factorise 8100

Answer 13

And it will always be true for any a and b that:
\[\sqrt{a*a*b}=\sqrt{a^2b}=\sqrt{a^2}\sqrt{b}=a\sqrt{b}\]And sometimes b itself will have more ways to factor it and you can work to simplify \[\sqrt{b}\]

Answer 14

tue for real factors yes

Answer 15

@sparky16 do you mean something like \[\sqrt{2 \times 2 \times 3 \times 3}\] if so pull out 2 and 3 \[\implies 2(3) \sqrt{\cancel{2 \times 2 \times 3 \times 3}} \]

Answer 16

\[\sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 3} \]
\[\implies 2\sqrt{\cancel{2 \times 2} \times 2 \times 2 \times 3 \times 3}\]
\[\implies 2(2) \sqrt{ \cancel{2 \times 2 \times 2 \times 2 \times 2} \times 3 \times 3}\]
\[\implies 2(2)(3)\sqrt{\cancel{2 \times 2\times 2\times 2\times 3 \times 3}}\]
does that help?

Answer 17

@igbasallote- like 2 x 2 x 2 x 2 x 3 x 3 ( just times them all together right?)

or do you take about the group of 2 and just time it by 3 to get 6 or is 12 because you times all of them?

Answer 18

uhh i think i exceeded by one 2 in the square root in the second line...iignore that typo

Answer 19

it's 12...you multiply everything you pulled out

Answer 20

okay thankyou and unklerhaukus.... when there are multiple pairs to pull out what do i do?

Answer 21

8100 has 5 pairs to pull out so what stays inside the radical?

Answer 22

\[\sqrt{8100}=\sqrt{81\times100}\]\[=\sqrt{9^2\times10^2}=\sqrt{9^2}\times\sqrt{10^2}\]\[=9\times10=90\]

Answer 23

i did the factor tree and i got 90 just like you thank you... and one last question... when dealing with the number inside the radical, do i just times together the leftover numbers that aren't pairs?

Answer 24

yes @sparky16 same thing i did in that tutorial i showed you

Answer 25

or you could have done it like this
\[\sqrt{8100}=\sqrt{2^2\times3^4\times5^2}=\sqrt{2^2\times(3^2)^2\times5^2}\]
\[=\sqrt{2^2}\times\sqrt{(3^2)^2}\times\sqrt{5^2}\]\[=2\times3^2\times5\]

Answer 26

thank you so much to everyone that helped! I understand it now! Next time I log on I may have another question in the future so look for my name cuz u were all amazing help!(:

Answer 27

haha tag us :p lol

Answer 28

alright will do when I need help! especially you igbasalote! thanks! kk bye