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Mathematics 19 Online
OpenStudy (coledrf):

help!

OpenStudy (coledrf):

simplify \[\sqrt{140}\]

OpenStudy (anonymous):

Alrighty, do you know what you can multiply to get 140?

OpenStudy (coledrf):

lol 70 and 2

OpenStudy (anonymous):

alrighty, so now you have this\[\sqrt{140}=\sqrt{2*70}\]break it down further, what are multiples of 70?

OpenStudy (coledrf):

7 and 10

OpenStudy (coledrf):

and multiples of 10 are 2 and 5

OpenStudy (anonymous):

alrighty, since you used those multiples, we need to break it down one more time, since we have\[\sqrt{140}=\sqrt{70*2}=\sqrt{7*10*2}\]

OpenStudy (anonymous):

there you go, now you understand so

OpenStudy (anonymous):

now we have this\[\sqrt{140}=\sqrt{2*2*5*7}=\sqrt{2}*\sqrt{2}*\sqrt{5}*\sqrt{7}\]so do you know what happens when you multiply\[\sqrt{2}*\sqrt{2}\]

OpenStudy (coledrf):

\[\sqrt{4}\]

OpenStudy (anonymous):

yup, and what does that simplify to?

OpenStudy (coledrf):

2

OpenStudy (anonymous):

Bam, so now look what we have,\[2*\sqrt{7}*\sqrt{5}=2\sqrt{35}\]sadly that's as far as you can simplify this, so you're done once you get that far

OpenStudy (coledrf):

yay thanks ! now what about this? what is the simplified form of \[\sqrt{48n^9}\]

OpenStudy (coledrf):

that is a 9

OpenStudy (anonymous):

n to the power of 9 right?

OpenStudy (coledrf):

yes

OpenStudy (anonymous):

Alrighty, so here are some rules for you to remember that will help, when asked to simplify a square root, try to take the regular number, and break it down to numbers that are perfect squares, so 48 should be broken down like this\[\sqrt{48}=\sqrt{4*12}=\sqrt{4*4*3}\]so you can simplify that, and here's the next rule\[n^9=n^4*n^4*n\]why is this true? because\[x^{a+b}=x^a*x^b\]so you can break up the power

OpenStudy (coledrf):

ok so the answer would be\[4n^4 \sqrt{3n}\]

OpenStudy (anonymous):

Great job! Yes, that's correct

OpenStudy (coledrf):

:) thanks

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