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How do you evaluate #\frac { 7} { 2+ 2\sqrt { 3} }#?

1 Answer

Shwetank Mauria

Dec 28, 2016

#7/(2+2sqrt3)=(7sqrt3-7)/4#

Explanation:

#7/(2+2sqrt3)=7/(2sqrt3+2)#

= #7/(2sqrt3+2)xx(2sqrt3-2)/(2sqrt3-2)# and using #(a+b)(a-b)=(a^2-b^2)#, we get

#(7(2sqrt3-2))/((2sqrt3)^2-2^2)#

= #(14sqrt3-14)/(12-4)#

= #(14sqrt3-14)/8# and dividing numerator and denominator by #2#

= #(7sqrt3-7)/4#

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