Square Root

Algebra Level 2

6 3 − 2 3 + 1 = ? \large \frac { 6 }{ \sqrt { 3 } } \quad -\quad \frac { 2 }{ \sqrt { 3 } \quad +\quad 1 } =\, ?

2 3 + 1 2\sqrt { 3\quad } +1 3 − 1 \sqrt { 3 } -1 3 \sqrt { 3 } 3 + 1 \sqrt { 3 } +1

Kaan Çamur by Kaan Çamur , 29, Turkey

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2 solutions

Mateus Gomes
Feb 8, 2016

6 3 − 2 3 + 1 = 6 ( 3 + 1 ) − 2 ( 3 ) ( 3 ) ( 3 + 1 ) = 4 3 + 6 ( 3 ) ( 3 + 1 ) = ( 3 ) ( 4 + 2 3 ) ( 3 ) ( 3 + 1 ) = ( 4 + 2 3 ) ( 3 + 1 ) = ( 3 + 1 ) 2 ( 3 + 1 ) = ( 3 + 1 ) \Large \frac { 6 }{ \sqrt { 3 } } \quad -\quad \frac { 2 }{ \sqrt { 3 } \quad +\quad 1 } =\frac{6(\sqrt3+1)-2(\sqrt3)}{(\sqrt3)(\sqrt3+1)}=\frac{4\sqrt3+6}{(\sqrt3)(\sqrt3+1)}=\frac{(\sqrt3)(4+2\sqrt3)}{(\sqrt3)(\sqrt3+1)}=\frac{(4+2\sqrt3)}{(\sqrt3+1)}=\frac{(\sqrt3+1)^2}{(\sqrt3+1)}=\Large\color{#3D99F6}{\boxed{(\sqrt3+1)}}

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Kay Xspre
Feb 9, 2016

Simplify 6 3 = 2 3 \displaystyle \frac{6}{\sqrt{3}} = 2\sqrt{3} and 2 3 + 1 = 2 ( 3 − 1 ) ( 3 ) 2 − 1 2 = 3 − 1 \displaystyle \frac{2}{\sqrt{3}+1} = \frac{2(\sqrt{3}-1)}{(\sqrt{3})^2-1^2} = \sqrt{3}-1 , therefore the solution is 3 + 1 \sqrt{3}+1

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