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

2 Answers

We have that

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

Jun 27, 2018

Answer:

Take the needed answer form with variables and solve for them

Explanation:

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)#