How do you simplify #sqrt(3-2*sqrt(2))#?

Question

Isaiah Anthony · Accepted Answer

We have that #sqrt(3-2*sqrt(2))=sqrt((sqrt2)^2-2sqrt2*1+1^2)=(sqrt((sqrt2-1))^2)=sqrt2-1#

Julia Adams · Answer

Take the needed answer form with variables and solve for them

The answer above is completely correct, but it may help the student to see how to reach it: We seek a square root of #3-2sqrt(2)#. This must have the form #a+bsqrt(2)#. #(a+bsqrt(2))^2=a^2+2ab sqrt(2)+2b^2# So #a^2+2b^2=3# and #2ab=-2rArrab=-1# These are two simultaneous equations for #a# and #b#. Rearrange the second: #b=-1/a# Substitute into the first: #a^2+2/a^2=3# #a^4+2=3a^2# #a^4-3a^2+2=0# Note that this is a quadratic in #a^2#: #(a^2)^2-3(a^2)+2=0# Factorise: #(a^2-2)(a^2-1)=0# This gives us two possible solutions for #a^2#: #2# and #1#, and so the four solutions for #a#: #+-sqrt(2)# and #+-1#. We are looking for integer solutions for #a#, and so #+-1# are possible solutions. But the other two are possible too - they can simply be folded in to the #sqrt(2)# term. This wouldn't have been possible if we'd had the root of some other number in the solution for #a#, but this solution is a special case. Now use the second equation to deduce the four equivalent solutions for #b#: #b=-1/a# #b=bar(+)1/sqrt(2)=bar(+)1/2sqrt(2)# and #bar(+)1#. So we have the four solution pairs #(a,b)#: #(sqrt(2),-1/2sqrt(2))# #(-sqrt(2),1/2sqrt(2))# #(1,-1)# #(-1,1)# This is a bit suspicious - we expect only two solutions, positive and negative square roots, so we wonder if some of these are identical to each other: When we substitute them in to our desired expression #a+bsqrt(2)#, we get: #sqrt(2)-1/2sqrt(2)sqrt(2)=-1+sqrt(2)# #-sqrt(2)+1/2sqrt(2)sqrt(2)=1-sqrt(2)# #1-sqrt(2)# #-1+sqrt(2)# So the two solutions with #sqrt(2)# are identical to the two simpler solutions, so we can get rid of them. We now have two solutions, positive and negative square roots: #1-sqrt(2)# #-1+sqrt(2)# When we take the written square root of a quantity, it is implied that the desired root is the positive root, the "principal value" of the square root function. So we take the single solution that comes from #a=+1#: #1-sqrt(2)# Double check: Make sure that this produces the desired answer: #(1-sqrt(2))^2=1-2sqrt(2)+2=3-2sqrt(2)#