Question 1107096
.
I am familiar with this problem . . . 


It was posted to the forum a week or two ago,  and I solved it . . . 


So, &nbsp;I know that &nbsp;<U>the formulation in &nbsp;THIS&nbsp; post IS &nbsp;&nbsp;NOT &nbsp;EXACTLY &nbsp;PRECISE</U>.


Therefore, &nbsp;I edited it, &nbsp;and the &nbsp;EDITED &nbsp;formulation &nbsp;(the only correct, &nbsp;valid and right version) &nbsp;is &nbsp;<U>THIS</U>:


<pre>
    Evaluate   {{{ sqrt( (7/3)+sqrt( (7/9)+sqrt( (7/3)+sqrt( (7/9) + ellipsis) ) ) ) }}} 
</pre>

Below is the <U>solution</U>:


<pre>
Let  us consider, for brewity of writing, more general expression

{{{sqrt(a + sqrt(b + sqrt(a + sqrt(b) + ellipsis)))}}} = x,    (1)

where  a = {{{7/3}}},  b = {{{7/9}}}.  Square (1)  (both sides).  You will get then  

{{{a + sqrt(b + sqrt(a + sqrt(b) + ellipsis)))}}} = {{{x^2}}},

{{{x^2-a}}} = {{{sqrt(b + sqrt(a + sqrt(b) + ellipsis)))}}}.        (2)


Square (2)   (both sides).  You will get then

{{{(x^2-a)^2}}} = {{{b+sqrt(a + sqrt(b) + ellipsis)))}}}.         (3)

{{{(x^2-a)^2-b}}} = {{{sqrt(a + sqrt(b) + ellipsis)))}}}.         (4)


Notice that the right side of the expression (4) is the same as the given expression,  so it is equal to x.  Thus you have 

{{{(x^2-a)^2 - b}}} = x.                  


It is equivalent to


{{{x^4 -2a*x^2 + a^2}}} - {{{b}}} = x,    or

{{{x^4 - 2a*x^2 - x + (a^2-b)}}} = 0.


Now substitute here  a = {{{7/3}}},  b = {{{7/9}}}. You will get this equation in the form

{{{x^4 - (14/3)*x^2 - x + 14/3}}} = 0,   or, multiplying all the terms by 3

{{{3x^4 -14x^2 - 3x + 14}}} = 0.


The plot of the last polynomial is shown below.



{{{graph( 330, 330, -1.5, 5.5, -10.5, 15.5,
          3x^4 - 14x^2 - 3x + 14
)}}}


Plot y = {{{3x^4 - 14x^2 - 3x + 14}}}



It clearly shows that  x= 1  and  x= 2 are the roots.  Having this HINT, you can check it MANUALLY  (as I did . . . ).

The two other roots of the polynomial are complex numbers.


Since  the value of  {{{sqrt(7/3 + sqrt(7/9 + sqrt(7/3 + sqrt(7/9) + ellipsis)))}}}  is, obviously, real number greater than 1, it can be only 2.


It proves that  {{{sqrt(7/3 + sqrt(7/9 + sqrt(7/3 + sqrt(7/9) + ellipsis)))}}} = 2.


<U>Answer</U>.  {{{sqrt(7/3 + sqrt(7/9 + sqrt(7/3 + sqrt(7/9) + ellipsis)))}}} = 2.


<U>Check</U>.   {{{sqrt(7/3 + sqrt(7/9 + sqrt(7/3 + sqrt(7/9))))}}} = 1.984 (approximately).
</pre>

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See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluating-expressions-that-contain-infinitely-many-square-roots.lesson>Evaluating expressions that contain infinitely many square roots</A>

in this site.