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 May 4th, 2014, 12:44 PM #1 Member   Joined: Apr 2014 From: Florida Posts: 69 Thanks: 4 square root squared I know how, but don't fully understand why, I must have missed it in my lessons. How do you get the square root squared $$(\sqrt{91})^2$$ I know the answer is 91; how do you arrive at this answer? To make it easy, let's use a smaller number $$(\sqrt{8})^2 = 8$$ Please explain how. I need to see the math so I understand how it's done. I would assume it is like so: $$(\sqrt{8})^2 = (\sqrt{4}\cdot \sqrt{2})^2 = (\sqrt{4})^2 \cdot (\sqrt{2})^2 = (\sqrt{4} \cdot \sqrt{4}) \cdot (\sqrt{2} \cdot \sqrt{2}) = \sqrt{16} \cdot \sqrt{4} = 4 \cdot 2 = 8$$ But I do not think this is correct. Something just feels wrong about it since this relies on splitting into 2 numbers, one being a perfect square. This can't be done with $$\sqrt{91}$$ Please work out both so I can see how its done with both the odd number and even number. Last edited by skipjack; May 5th, 2014 at 09:20 PM. May 4th, 2014, 01:00 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Sqrt(4) = 2 ; 2squared = 4 ; so SQRT(4)squared = 4 ... Thanks from agentredlum May 4th, 2014, 01:08 PM #3 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,948 Thanks: 1139 Math Focus: Elementary mathematics and beyond $\displaystyle (\sqrt{91})^2=\sqrt{91}\cdot\sqrt{91}=\sqrt{8281}= 91$ Thanks from agentredlum May 4th, 2014, 02:36 PM #4 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 √n is BY DEFINITION the number which when squared = n. May 4th, 2014, 03:21 PM   #5
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Quote:
 Originally Posted by greg1313 $\displaystyle (\sqrt{91})^2=\sqrt{91}\cdot\sqrt{91}=\sqrt{8281}= 91$
Thank you. This explained it.

Quote:
 Originally Posted by mathman √n is BY DEFINITION the number which when squared = n.
True, but how is it proven?
$$(\sqrt{8})^2 = \sqrt{8} \cdot \sqrt{8} = \sqrt{64} = 8$$

So my method is incorrect $$(\sqrt{8})^2 = (\sqrt{4}\cdot \sqrt{2})^2 = (\sqrt{4})^2 \cdot (\sqrt{2})^2 = (\sqrt{4} \cdot \sqrt{4}) \cdot (\sqrt{2} \cdot \sqrt{2}) = \sqrt{16} \cdot \sqrt{4} = 4 \cdot 2 = 8$$ even though the result was the same, because it couldnt be used for $$(\sqrt{91})^2$$ it would be incorrect. May 4th, 2014, 11:19 PM #6 Math Team   Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Everything you did is correct (but longer). You can do $(\sqrt{91})^2$ your way also $$(\sqrt{91})^2 = ( \sqrt{13} \cdot \sqrt{7})^2 = ( \sqrt{13} )^2 \cdot (\sqrt{7})^2 = \\ \\ ( \sqrt{13} \cdot \sqrt{13} ) \cdot ( \sqrt{7} \cdot \sqrt{7}) = \sqrt{169} \cdot \sqrt{49} = 13 \cdot 7 = 91$$ Thanks from kevintampa5 May 5th, 2014, 03:45 AM #7 Senior Member   Joined: Apr 2014 From: UK Posts: 921 Thanks: 331 root(91) can also be written as 91^0.5 So, (91^0.5)^2 = 91^(0.5*2) = 91 Thanks from agentredlum May 5th, 2014, 12:45 PM #8 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 √n is BY DEFINITION the number which when squared = n. Definitions don't need to be proved. Thanks from agentredlum May 5th, 2014, 04:15 PM   #9
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Quote:
 Originally Posted by mathman √n is BY DEFINITION the number which when squared = n. Definitions don't need to be proved.
Humor me. Whether or not it need be done, I asked for it to be proved so I understand why the definition is so. I do appreciate your input, however. May 5th, 2014, 04:16 PM   #10
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 Originally Posted by agentredlum Everything you did is correct (but longer). You can do $(\sqrt{91})^2$ your way also $$(\sqrt{91})^2 = ( \sqrt{13} \cdot \sqrt{7})^2 = ( \sqrt{13} )^2 \cdot (\sqrt{7})^2 = \\ \\ ( \sqrt{13} \cdot \sqrt{13} ) \cdot ( \sqrt{7} \cdot \sqrt{7}) = \sqrt{169} \cdot \sqrt{49} = 13 \cdot 7 = 91$$ That is good to know. Now I understand it fully knowing it can be done that way as well. Tags root, square, squared Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post TwoTwo Algebra 43 December 1st, 2013 02:08 AM candi678 Number Theory 2 January 28th, 2010 03:31 PM cadman12 Advanced Statistics 1 March 27th, 2008 02:53 AM jared_4391 Algebra 3 August 8th, 2007 09:06 AM rain Academic Guidance 1 December 31st, 1969 04:00 PM

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