Question 1199848
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{{{sqrt(8*sqrt(11)+27) = a+b*sqrt(c)}}} where a,b,c are integers and {{{c >= 0}}}


{{{(sqrt(8*sqrt(11)+27))^2 = (a+b*sqrt(c))^2}}}


{{{8*sqrt(11)+27 = a^2+2ab*sqrt(c)+b^2*c}}}


{{{27+8*sqrt(11) = (a^2+b^2*c)+2ab*sqrt(c)}}}


Equate corresponding terms to form this system of equations
{{{system(27 = a^2+b^2c,8 = 2ab,11 = c)}}}
Use c = 11 and we get this slightly reduced system
{{{system(27 = a^2+11b^2,8 = 2ab)}}}
which turns into
{{{system(27 = a^2+11b^2,4 = ab)}}}
Through a bit of trial and error, or the substitution method, you should find these solutions
a = 4
b = 1


So,
{{{sqrt(8*sqrt(11)+27) = a+b*sqrt(c)}}}


{{{sqrt(8*sqrt(11)+27) = 4+1*sqrt(11)}}}


{{{sqrt(8*sqrt(11)+27) = 4+sqrt(11)}}} ... can be verified using WolframAlpha


and
{{{sqrt(8*sqrt(11)+27) = y-sqrt(x-2y)}}}


{{{4+sqrt(11) = y-sqrt(x-2y)}}}


At first things seem fine.
The 4 and y match up to show y = 4
But the square root terms do not match up fully.
The left hand side (LHS) has a positive square root, while the right hand side (RHS) has the square root as negative. 
This is a contradiction that you'll need to ask your teacher about for further clarification.


If we ignored the positive and negative discrepancy, then we can equate radicands, plug in y = 4 and solve for x.
11 = x-2y
11 = x-2*4
11 = x-8
x = 11-8
x = 19


But as mentioned before, we run into a contradiction
{{{4+sqrt(11) = y-sqrt(x-2y)}}}
{{{4+sqrt(11) = 4-sqrt(19-2*4)}}}
{{{4+sqrt(11) = 4-sqrt(11)}}}
{{{7.316625 = 0.683375}}}


It's possible your teacher made a typo somewhere.
I would ask them for clarification.


The tutor Edwin McCravy reached a similar conclusion in the link below.
<a href = "https://www.algebra.com/algebra/homework/Radicals/Radicals.faq.question.1151181.html">https://www.algebra.com/algebra/homework/Radicals/Radicals.faq.question.1151181.html</a>
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