Question 681800
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I'm trying to prove that

{{{sqrt(4 - 2 * sqrt(3)) + sqrt(12 - 6 * sqrt(3)) = 2}}}            
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This author's proof is different from the one the other person who responded, provided. His method
may not be what the person who needs help, is looking for. Then again, this hunch may be WRONG!

Having said that, let's focus on the L.H.S. because that's what's needed to be proven to equal 2.
                        {{{sqrt(4 - 2sqrt(3))}}} + {{{sqrt(12 - 6sqrt(3))}}}
                        {{{sqrt(4 - 2sqrt(3))}}} + {{{sqrt(12 - 2(3)sqrt(3))}}} ---- Changing 6 to 2(3)
                        {{{sqrt(4 - 2sqrt(3))}}} + {{{sqrt(12 - 2sqrt(9)sqrt(3))}}} ------- Changing 3 to {{{sqrt(9)}}} 
               {{{sqrt(3 + 1 - 2sqrt(3)sqrt(1))}}} + {{{sqrt(9 + 3 - 2sqrt(9)sqrt(3))}}} ----- Changing 4 to 3 + 1, 3 to {{{sqrt(3)sqrt(1)}}}, and 12 to 9 + 3                 
{{{sqrt((sqrt(3))^2 + (sqrt(1))^2 - 2sqrt(3)sqrt(1))}}} + {{{sqrt((sqrt(9))^2 + (sqrt(3))^2 - 2sqrt(9)sqrt(3))}}} ------ Converting {{{system(matrix(4,3, 3, to, (sqrt(3))^2, 1, to, (sqrt(1))^2, 9, to, (sqrt(9))^2, 3, to, (sqrt(3))^2))}}}
The above is in the form:
                          {{{(a[1] - b[1])^2}}}  + {{{(a[2] - b[2])^2}}}, with {{{system(matrix(4,3, a[1], being, sqrt(3), b[1], being, sqrt(1), a[2], being, sqrt(9), b[2], being, sqrt(3)))}}}, and so:
{{{sqrt((sqrt(3))^2 + (sqrt(1))^2 - 2sqrt(3)sqrt(1))}}} + {{{sqrt((sqrt(9))^2 + (sqrt(3))^2 - 2sqrt(9)sqrt(3))}}} then becomes:
                   {{{sqrt((sqrt(3) - sqrt(1))^2)}}} +  {{{sqrt((sqrt(9) - sqrt(3))^2)}}}
                         {{{sqrt(3) - sqrt(1)}}}    +    {{{sqrt(9) - sqrt(3)}}} ----- Cancelling SQUARE and SQUARE ROOT
                             {{{sqrt(3) - 1}}} + {{{3 - sqrt(3) = highlight(2))}}}QED!</font></font></font></b></pre>