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Mathematics 7 Online
OpenStudy (calculusxy):

Can someone explain to me how to find the square root of 50?

OpenStudy (calculusxy):

@TuringTest

OpenStudy (anonymous):

Are you trying to get an approximation?

OpenStudy (calculusxy):

No I want to simplify it into another form of square root.

jimthompson5910 (jim_thompson5910):

can you factor 50 where one factor is a perfect square?

OpenStudy (calculusxy):

Yes 25.

jimthompson5910 (jim_thompson5910):

25 times what?

OpenStudy (calculusxy):

2

jimthompson5910 (jim_thompson5910):

so, \[\Large \sqrt{50} = \sqrt{25*2}\] \[\Large \sqrt{50} = \sqrt{25}*\sqrt{2}\] \[\Large \sqrt{50} = 5\sqrt{2}\] in step 2, I'm using rule #1 from this page (look under the "distributing" section) http://www.mathwords.com/s/square_root_rules.htm

OpenStudy (calculusxy):

Can you explain to me why you have kept the 5 outside. I understood that the square root of twenty five is five. But can you emphasize this to me a bit moreover?

jimthompson5910 (jim_thompson5910):

well I broke up \(\Large \sqrt{25*2}\) into \(\Large \sqrt{25}*\sqrt{2}\) using that rule on the given link

jimthompson5910 (jim_thompson5910):

then I replaced all of \(\Large \sqrt{25}\) with 5 since \(\Large \sqrt{25}=5\)

OpenStudy (calculusxy):

But why have you kept the five outside and the two inde the radical sign?

OpenStudy (calculusxy):

*inside

jimthompson5910 (jim_thompson5910):

if 5 remained inside the root, then you'd have \(\Large \sqrt{5}\) but \(\Large \sqrt{25} \neq \sqrt{5}\)

jimthompson5910 (jim_thompson5910):

the 2 remains in the root sign because the square root of 2 is some irrational number

OpenStudy (calculusxy):

O ya thanks.

jimthompson5910 (jim_thompson5910):

you're welcome

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