Question 1030250
<pre>I need help simplifying this expression:
{{{sqrt(24+8sqrt(5))}}}
I removed the common factor out of the square root to obtain {{{2sqrt(6+2sqrt(5))}}}, but the answer key says it is {{{2+2sqrt(5)}}}.
How is it possible? Am I missing out on a rule here?
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This happens to be one of those RARE cases when you can actually factor out a COMMON factor, which happens to be a
PERFECT SQUARE (in 4). This is what you did:
{{{sqrt(24 + 8sqrt(5))}}} = {{{sqrt(4(6 + 2sqrt(5)))}}} 
                            {{{sqrt(4) * sqrt(6 + 2sqrt(5))}}} ----- Applying {{{sqrt(m*n)}}} = {{{sqrt(m)sqrt(n)}}}         
                               {{{2sqrt(6 + 2sqrt(5))}}}....At this point, you haven't fully simplified the
                                                       expression, and should've continued as follows:  
                            
                   {{{2sqrt(5 + 1 + 2sqrt(5*1))}}} ---- Changing 6 to 5 + 1, and 5 to 5*1  
                    {{{2sqrt(5 + 1 + 2sqrt(5)sqrt(1))}}} ---- Applying {{{sqrt(m*n)}}} = {{{sqrt(m)sqrt(n)}}}         
   {{{2sqrt((sqrt(5))^2 + (sqrt(1))^2 + 2sqrt(5)sqrt(1))}}} --- Converting {{{system(matrix(2,3, 5, to, (sqrt(5))^2, 1, to, (sqrt(1))^2))}}}
The above is in the form: {{{(a + b)^2}}}, with {{{system(matrix(2,3, a, being, sqrt(5), b, being, sqrt(1)))}}}, and so:
{{{2sqrt((sqrt(5))^2 + (sqrt(1))^2 + 2sqrt(5)sqrt(1))}}} then becomes: {{{2sqrt((sqrt(5) + sqrt(1))^2)}}} 
                                                                                {{{2(sqrt(5) + sqrt(1))}}} ----- Cancelling SQUARE and SQUARE ROOT
                                                                                {{{2(sqrt(5) + 1)}}} = {{{highlight(2sqrt(5) + 2)}}}
                                    This certainly matches the answer key: {{{2+2sqrt(5)}}}
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On the other hand, this author would've SIMPLIFIED the SURD from the onset, as follows:
                                 {{{sqrt(24 + 8sqrt(5))}}}
                       {{{sqrt(24 + 2(4)sqrt(5))}}} ----- Replacing 8 with factors, 2 & 4
                         {{{sqrt(24 + 2sqrt(16)sqrt(5))}}} ----- Converting 4 to {{{sqrt(16)}}}
                              {{{sqrt(24 + 2sqrt(80))}}} ----- Applying {{{sqrt(m)sqrt(n)}}} = {{{sqrt(mn))}}}
                        {{{sqrt(24 + 2sqrt(20*4))}}}                
                    {{{sqrt(20 + 4 + 2sqrt(20)sqrt(4))}}} ---- Changing 24 to 20 + 4, and applying {{{sqrt(m*n)}}} = {{{sqrt(m)sqrt(n)}}}           
  {{{sqrt((sqrt(20))^2 + (sqrt(4))^2 + 2sqrt(20)sqrt(4))}}} ---- Converting {{{system(matrix(2,3, 20, to, (sqrt(20))^2, 4, to, (sqrt(4))^2))}}}
The above is in the form: {{{(a + b)^2}}}, with {{{system(matrix(2,3, a, being, sqrt(20), b, being, sqrt(4)))}}}, and so:
    {{{sqrt((sqrt(20))^2 + (sqrt(4))^2 + 2sqrt(20)sqrt(4))}}} then becomes:      {{{sqrt((sqrt(20) + sqrt(4))^2)}}}
                                                                                                 {{{sqrt(20) + sqrt(4)}}} ----- Cancelling SQUARE and SQUARE ROOT
                                                                                                 {{{highlight(2sqrt(5) + 2)}}}
                              This certainly matches the answer key: {{{2+2sqrt(5)}}}</font></font></font></b></pre>