How do you simplify #(5 + sqrt 3) /( 2 - sqrt3)#?

1 Answer
Tony B
Apr 11, 2016

#(5+sqrt(3))/(2-sqrt(3))=13+7sqrt(3)#

Explanation:

Using #a^2-b^2=(a+b)(a-b)#

You can write the value 1 in many ways: #2/2; 16/16; or 1=(2+sqrt(3))/(2+sqrt(3))#

Multiply by 1 but in the form #1=(2+sqrt(3))/(2+sqrt(3))# giving

#(5+sqrt(3))/(2-sqrt(3)) xx (2+sqrt(3))/(2+sqrt(3))" "=" "( (5+sqrt(3))(2+sqrt(3)))/(2^2 -[(sqrt(3))^2])#

#=> (10+5sqrt(3)+2sqrt(3)+3)/(4-3)#

#=>13+7sqrt(3)#

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