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