Question 1118304
x=sqrt(3+2sqrt(2)) + sqrt(3-2sqrt(2))
A) fund the exact value of x^2
B) hence, determine the simplified form of x
<pre>
{{{x=sqrt(3+2sqrt(2)) + sqrt(3-2sqrt(2))}}}

To make things easier, let

{{{sqrt(3+2sqrt(2))=A}}} and {{{sqrt(3-2sqrt(2))=B}}}

Then 

{{{x=A + B}}}

{{{x^2 = (A+B)^2}}}

{{{x^2 = (A+B)(A+B)}}}

{{{x^2=A^2+AB+AB+B^2}}}

{{{x^2=A^2+2AB+B^2}}}

Now since 

{{{x^2 = A^2+2AB+B^2}}}, we'll need AČ, BČ, and AB

{{{A^2 = (sqrt(3+2sqrt(2)))^2}}}

{{{A^2 = 3+2sqrt(2)}}}

{{{B^2 = (sqrt(3-2sqrt(2)))^2}}}

{{{B^2 = 3-2sqrt(2)}}}

{{{AB = (sqrt(3+2sqrt(2)))(sqrt(3-2sqrt(2)))))}}}

{{{AB = sqrt((3+2sqrt(2))(3-2sqrt(2)))}}}

{{{AB = sqrt((3+2sqrt(2))(3-2sqrt(2)))}}}

{{{AB = sqrt(9-6sqrt(2)+6sqrt(2)-4*2))}}}

{{{AB = sqrt(9-4*2)}}}

{{{AB = sqrt(9-8)}}}

{{{AB = sqrt(1)}}}

{{{AB = 1}}}

Substituting in

{{{x^2 = A^2+2AB+B^2}}},

{{{x^2 = (3+2sqrt(2))+2(1)+(3-2sqrt(2))}}},

{{{x^2 = 3+2sqrt(2)+2+3-2sqrt(2)}}},

{{{x^2 = 3+2+3}}},

{{{x^2=8}}}

{{{x="" +- sqrt(8)}}}

But x cannot be negative.

{{{x=sqrt(4*2)}}}

{{{x=sqrt(4)sqrt(2)}}}

{{{x=2sqrt(2)}}}

Edwin</pre>