Question 1046294
.
Kindly help me solving it 
What are the roots obtained by solving equation 
{{{z^2-6z+9=4 (z^2-6z+6)^(1/2)}}}?
Well i tried hard and simplified it upto
{{{z^4-12z^3+38z^2-12z+63=0}}}
But i can't go further, so please guide me.
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<pre>
{{{z^2-6z+9}}} = {{{4*sqrt(z^2-6z+6)}}}

Rewrite equivalently

{{{(z-3)^2}}} = {{{4*sqrt((z-3)^2-3)}}}.

Square both sides:

{{{(z-3)^4}}} = {{{16*((z-3)^2-3)}}}.

Introduce new variable x = {{{(z-3)^2}}}. The last equation takes the form

{{{x^2}}} = {{{16(x -3)}}},  or

{{{x^2 - 16x + 48}}} = {{{0}}}.

Solve by any way (factoring or the quadratic formula). You will get the roots

  x = 12  and/or  x = 4.

Thus you have two options:


1.  {z-3)^2 = 4   --->  z-3 = 2  or  z-3 = -2  --->  z = 5  or  z = 1.


2.  (z-3)^2 = 12  --->  z-3 = +/-{{{sqrt(12)}}} = +/-{{{2*sqrt(3)}}}  --->  z = {{{3+2sqrt(3)}}}  and/or  z = {{{3-2sqrt(3)}}}.


Check that all these roots are the solutions of the original equation.


<U>Answer</U>. The solutions are  z = 1,  z = 5,  z = {{{3+2sqrt(3)}}},  z = {{{3-2sqrt(3)}}}.
</pre>

<TABLE> 
  <TR>
  <TD> 

{{{graph( 330, 330, -2.5, 8.5, -2.5, 15.5,
          x^2-6x+9, 4*sqrt(x^2-6x+6)
)}}}


Plots y = {{{x^2-6x+9}}} (red), y = {{{4*sqrt(x^2-6x+6)}}} (green)

  </TD>
  </TR>
</TABLE>