How do you simplify #(4+sqrt2 )div (8 - sqrt2)#?

1 Answer
Sep 20, 2015

Answer:

#(17+6sqrt2)/31#

Explanation:

Let's rewrite this in the fractional form:
#(4+sqrt2)/(8-sqrt2)#

Then, I would multiply, top and bottom, by #(8+sqrt2)# so that:
#(4+sqrt2)/(8-sqrt2) * (8+sqrt2)/(8+sqrt2)#

(Note that #(8+sqrt2)/(8+sqrt2)=1#, so multiplying by this factor does not change the results at all).

I chose #(8+sqrt2)# because
#(a-b)*(a+b)=a^2-b^2#
and I am hoping it will make the solution easier.

Then the fraction becomes:
#((4+sqrt2)(8+sqrt2))/(8^2 - (sqrt2)^2) = (32+12sqrt2+2)/(64-2)#
which is
#=(34+12sqrt2)/(62)#
We can factor out 2 from the top and bottom:
#=(2*(17+6sqrt2))/(2*31)#
#=(17+6sqrt2)/31#