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pennstate99
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Help with a square root 14 Sep 2005, 20:31
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Hi everyone...this is my first post. I'm totally stuck on how

24/(sqrt(2)) can be reduced to 12(sqrt(2)). At first I thought the book was wrong but I plugged it into excel and they are equal. What am I missing here?

Many thanks!



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vikramm
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Help with a square root 14 Sep 2005, 20:34
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24/sqrt(2) = 24*sqrt(2)/(sqrt(2)*sqrt(2)) (multiply/divide by sqrt(2))
= 24*sqrt(2)/2
=12*sqrt(2)
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chet719
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Help with a square root 15 Sep 2005, 16:25
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Or you could multiply top and bottom by -Sqrt(2)
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Folaa3
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Help with a square root 15 Sep 2005, 19:05
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pennstate99
Hi everyone...this is my first post. I'm totally stuck on how

24/(sqrt(2)) can be reduced to 12(sqrt(2)). At first I thought the book was wrong but I plugged it into excel and they are equal. What am I missing here?

Many thanks!
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Hi pennstate99, for more information about reducing irrational equations, you might want to read up on conjugates, rationalizing denominators and rationalizing numerators. I hope that helps
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pennstate99
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Help with a square root 15 Sep 2005, 19:17
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thanks guys. seems so obvious now :)
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Antmavel
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Help with a square root 15 Sep 2005, 19:27
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vikramm
24/sqrt(2) = 24*sqrt(2)/(sqrt(2)*sqrt(2)) (multiply/divide by sqrt(2))
= 24*sqrt(2)/2
=12*sqrt(2)
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You can also consider the following:
24=2*12=2^1*12^1
sqrt(2)=2^(1/2)
so 24/sqrt(2) = (2^1*12^1) / 2^(1/2) = 2^(1/2)*12^1 = sqrt(2)*12

:wink:
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richardj
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Help with a square root 15 Sep 2005, 23:34
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Or even just square the whole thing

24/sqrt(2) = sqrt(24*24)/sqrt(2)
= sqrt ((24*24)/2)
= sqrt((2*12)*(2*12)/2)
= sqrt(2*2*12*12/2)
= sqrt ((2*2)/2*(12*12))
= sqrt (2 * (12*12))
= sqrt (2) * sqrt (12*12)
= 12 * sqrt(2)



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!