Root squared problem i

Algebra Level 2

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B D A C

Heder Oliveira Dias by Heder Oliveira Dias , 38, Brazil

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3 solutions

Sanchit Batra
Jul 22, 2014

1/2*4=2 thus= (1+√1+1)^2 1^2+(√2)^2+2√2 =3+2√2

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can any one explain this in a better manner

Varun Raj - 6 years, 10 months ago

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Done .. Check my solution :)

Ahmad Hesham - 6 years, 10 months ago

( 1 + 1 + 1 2 2 2 ) 4 = ( 1 + 1 + 1 2 ) 2 \color{forestgreen}{\left(\sqrt[2]{1+\sqrt[2]{1+\sqrt[2]{1}}}\right)^4=(1+\sqrt[2]{1+1})^2} ( 1 + 2 2 ) 2 = 1 + 2 2 2 + 2 = 3 + 2 2 \color{#3D99F6}{(1+\sqrt[2]{2})^2=1+2\sqrt[2]{2}+2=3+\sqrt[2]{2}} So C \boxed{C} is correct.

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Ahmad Hesham
Jul 30, 2014

( 1 + 1 + 1 2 2 2 ) 4 = ( 1 + 1 + 1 2 2 ) 2 = ( 1 + 1 + 1 2 ) 2 = ( 1 + 2 2 ) 2 = 1 2 + 2 2 2 + ( 2 2 ) 2 = 1 + 2 2 2 + 2 = 3 + 2 2 2 (\sqrt [ 2 ]{ 1+\sqrt [ 2 ]{ 1+\sqrt [ 2 ]{ 1 } } } )^{ 4 }\\ \\ ={ (1+\sqrt [ 2 ]{ 1+\sqrt [ 2 ]{ 1 } } ) }^{ 2 }\\ \\ ={ (1+\sqrt [ 2 ]{ 1+1 } ) }^{ 2 }\\ \\ ={ { (1+\sqrt [ 2 ]{ 2 } ) } }^{ 2 }\\ \\ ={ 1 }^{ 2 }+2\sqrt [ 2 ]{ 2 } +{ (\sqrt [ 2 ]{ 2 } ) }^{ 2 }\\ \\ =1+2\sqrt [ 2 ]{ 2 } +2\\ \\ \boxed { =3+2\sqrt [ 2 ]{ 2 } }

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