TeemoTheTerific
Simplify
(2^1/2  2^1/2)^2
ill write it neater. Please help



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TeemoTheTerific
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dw:1359872594130:dw

TeemoTheTerific
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i know formula a^2  2ab + b^2

TeemoTheTerific
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i dont get it :P

TeemoTheTerific
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dw:1359872887253:dwbtw original question is

TeemoTheTerific
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thats original quesiton

hartnn
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dw:1359872899922:dw

TeemoTheTerific
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yes your right now hartnn :)

hartnn
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dw:1359872998532:dw

hartnn
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can you simplify that now ?

TeemoTheTerific
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answer 5?

TeemoTheTerific
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is there a way to do this question using indices laws?

TeemoTheTerific
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cause my teacher prefers that :P

hartnn
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yes, 5 is correct.

hartnn
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i have used law od indices
\(2^{1/2}.2^{1/2}= 2^{1/2+1/2}=2^1=2\)

hartnn
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dw:1359873234483:dw

hartnn
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dw:1359873310197:dw

hartnn
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got that^ ?

TeemoTheTerific
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yes thank you so much

hartnn
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welcome ^_^

whpalmer4
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\[(2^{1/2}  2^{1/2})^2 = (2^{1/2}2^{1/2})(2^{1/2}2^{1/2})=\]\[2^{1/2}*2^{1/2}2^{1/2}2^{1/2}  2^{1/2}2^{1/2} + 2^{1/2}2^{1/2} = \]\[2^12^02^0+2^{1} = 2  1  1 + \frac{1}{2} = \frac{1}{2}\]

whpalmer4
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Maybe easier to write \(a = 2^{1/2}\) and \(b = a^{1/2}\) then
\[(ab)^2 = a^2  2ab + b^2 = (2^{1/2})^{1/2}  2*2^{1/2}*2^{1/2} + (2^{1/2})^2 =\]\[2^1  2*2^0 + 2^{1} = 2^{1} = \frac{1}{2}\] and use some different properties :)