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Square Roots and Factoring

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by tonebeeze » Fri Mar 11, 2011 10:58 am
The value of √(2^5 * 3^2 * 5^4) is between which of the following pairs of numbers?

a. 0 and 100
b. 100 and 200
c. 200 and 300
d. 300 and 400
e. 400 and 500

OA = E
Last edited by tonebeeze on Fri Mar 11, 2011 11:33 am, edited 1 time in total.
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by anshumishra » Fri Mar 11, 2011 11:19 am
tonebeeze wrote:The value of √(2^5 * 3^2 * 5^2) is between which of the following pairs of numbers?

a. 0 and 100
b. 100 and 200
c. 200 and 300
d. 300 and 400
e. 400 and 500

OA = E
√(2^5 * 3^2 * 5^2) = 4*3*5*√2 = 60√2 ~= 84
So, ans. should be A.
Thanks
Anshu

(Every mistake is a lesson learned )
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by tonebeeze » Fri Mar 11, 2011 11:32 am
The square root sign is meant to cover all the terms under the radical, so 2^5 * 3^2 * 5^2 are intended to be under the radical. It is really hard to write out equations on here. Thus the square root of 180,000 should be between 400-500. So the answer is E
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by tonebeeze » Fri Mar 11, 2011 11:33 am
Anu, you were correct...I wrote the problem out incorrectly initially. I wrote 5^2 instead of 5^4. I apologize. I edited the equation in the initial post:
√(2^5 * 3^2 *5^4)
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by anshumishra » Fri Mar 11, 2011 11:36 am
tonebeeze wrote:Anu, you were correct...I wrote the problem out incorrectly initially. I wrote 5^2 instead of 5^4. I apologize. I edited the equation in the initial post:
√(2^5 * 3^2 *5^4)
No problem, it happens. Glad that you got the answer.
Thanks
Anshu

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