My Math Forum square root squared

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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: 14,143 Thanks: 1003 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: 7,923 Thanks: 1123 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: 6,710 Thanks: 675 √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: 898 Thanks: 329 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: 6,710 Thanks: 675 √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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Quote:
 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.

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