<PRE><B>Question 1124474</B>
&lt;br&gt;
The solution by tutor @mathlover is correct.&lt;br&gt;
But you might wonder how she came up with the idea of separating the 17 into 9+8, or the other things she did.&lt;br&gt;
What if you encounter another problem like this?  Is there a general method for evaluating an expression like this?&lt;br&gt;
Yes; there is.  Let me demonstrate.&lt;br&gt;
Consider an expression like&lt;br&gt;
{{{(sqrt(a)+sqrt(b))^2}}}&lt;br&gt;
FOILing gives us the expanded form&lt;br&gt;
{{{a + 2*sqrt(ab)+ b}}}
or
{{{(a+b)+2*sqrt(ab)}}}&lt;br&gt;
The type of problem you are working is where you are given an expression of this form and you are supposed to find the square root.&lt;br&gt;
The key to doing that is to get the expression you are given in EXACTLY the form shown.  If you do that, then your objective is to find two numbers whose sum is the whole number part of the expression (a+b) and whose product is the radicand (ab).&lt;br&gt;
A couple of simple examples:&lt;br&gt;
(1) {{{sqrt(8+2*sqrt(15))}}}
5+3=8; 5*3=15.  So {{{sqrt(8+2*sqrt(15)) = sqrt(5)+sqrt(3)}}}&lt;br&gt;
(2) {{{sqrt(16+2*sqrt(55))}}}
5+11=16; 5*11=55.  So {{{sqrt(16+2*sqrt(55)) = sqrt(5)+sqrt(11)}}}&lt;br&gt;
You might have to do some manipulation to get the given expression in the required form:&lt;br&gt;
(3) {{{sqrt(16+4*sqrt(15))}}}
The expression is not in the required form; the irrational part of the expression must be 2*sqrt(ab).  So modify the expression to make it that way:
{{{sqrt(16+2*sqrt(60))}}} (4 = 2*2; put the extra factor of 2 back under the radical as sqrt(4))
Now 10+6=16; 10*6=60.  So {{{sqrt(16+4*sqrt(15)) = sqrt(10)+sqrt(6)}}}&lt;br&gt;
Now we can look at solving your example using this method.&lt;br&gt;
{{{sqrt(17-12*sqrt(2))}}}
{{{sqrt(17-2*sqrt(72))}}} (12 = 2*6; put the extra factor of 6 back under the radical as sqrt(36))
8+9=17; 8*9=72; so&lt;br&gt;
{{{sqrt(17-12*sqrt(2)) = sqrt(9)-sqrt(8)}}}&lt;br&gt;
Note that with my examples the order of the two square roots did not matter; for example, sqrt(5)+sqrt(3) is the same as sqrt(3)+sqrt(5).&lt;br&gt;
However, with your example, since a square root is never negative, the answer has to be written as sqrt(9)-sqrt(8), because sqrt(8)-sqrt(9) would be negative.&lt;br&gt;
Here is one more example for you to try:&lt;br&gt;
{{{sqrt(36-12*sqrt(5))}}}&lt;br&gt;
See if you can get the answer, which is sqrt(30)-sqrt(6).</PRE>