How do you rationalize the denominator and simplify #(5+ sqrt 5)/(8- sqrt 5)#?

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Answer 1 · 2015-04-10T17:10:16

Multiply the numerator an denominator by the conjugate of the denominator, then look for simplification factors

#(5+sqrt(5))/(8-sqrt(5))#

#=(5+sqrt(5))/(8-sqrt(5)) * (8+sqrt(5))/(8+sqrt(5))#

#= (40 +13sqrt(5) +5)/(64-5)#

#= (45+13sqrt(5))/59#

Answer 2 · 2015-04-10T17:14:27

Multiply the fraction by #1# in the form #(8+sqrt5)/(8+sqrt5)#

#((5+sqrt5))/((8-sqrt5)) ((8+sqrt5))/((8+sqrt5))=(40+5sqrt5+8sqrt5+5)/(64-5)#

#=(45 + 13 sqrt5)/59#.

This works because of the product: #(a-b)(a+b)=a^2-b^2#.

If one or both ob #a#, #b# have square roots, then the product doesn't: for example: #(a-sqrtc)(a+sqrtc)=a^2 - (sqrtc)^2 = a^2-c#.