#4/(2+root2 3+root 2 7)# rationalize ?

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Answer 1 · 2018-05-06T15:10:09

#4/(2+sqrt3+sqrt7)=1+2/3sqrt3-1/3sqrt21#

By rationalizing we mean rationalizing the denominator. This is done as follows:

#4/(2+sqrt3+sqrt7)#

= #4/(2+sqrt3+sqrt7)xx(2+sqrt3-sqrt7)/(2+sqrt3-sqrt7)#

= #(8+4sqrt3-4sqrt7)/((2+sqrt3)^2-(sqrt7)^2)#

= #(8+4sqrt3-4sqrt7)/(4+4sqrt3+3-7)#

= #(8+4sqrt3-4sqrt7)/(4sqrt3)xxsqrt3/sqrt3#

= #(8sqrt3+4*3-4sqrt21)/(4*3)#

= #(8sqrt3+12-4sqrt21)/12#

= #1+2/3sqrt3-1/3sqrt21#