Find it!

$\sqrt{34 + 24\sqrt{2}}\\ = \sqrt {16 + 18 + 24\sqrt{2}}\\ = \sqrt {{(4)}^2 + {(3\sqrt{2})}^2 + (2 × 4 × 3\sqrt{2})}\\ = \sqrt {{(4 + 3\sqrt{2})}^2}\\ = 4 + 3\sqrt{2}$

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$\therefore$ The answer is $4 + 3 + 2 = \boxed{9}$ .

If you cannot see the perfect square, you can do it step by step.

Now, let $\sqrt{34+24\sqrt{2}} = p + q$ where $p = a$ and $q = b\sqrt{c}$

Square it:

$34+24\sqrt{2} = p^2 + q^2 + 2pq$

Since $q$ contains a square root, we know that:

$24\sqrt{2} = 2pq$ and $34 = p^2 + q^2$

Square the first equation:

$576(2) = 4p^2q^2\\ p^2q^2 = 288\\ q^2 = \dfrac{288}{p^2}$

Substitute this into the second equation:

$34 = p^2 + \dfrac{72}{p^2}\\ p^4 - 34p^2 + 288 = 0\\ (p^2 - 16)(p^2 - 18) = 0\\ p^2 = 16, 18\\ p = \pm 4, \pm \sqrt{18}$

Now, we know that $p=a$ is a positive integer, therefore the only possible value of $p$ is $4$

Substitute $p=4$ to find $q$ :

$24\sqrt{2} = 2(4)(q)\\ q=3\sqrt{2}$

Therefore, $\sqrt{34 + 24\sqrt{2}} = 4 + 3\sqrt{2}$

$a=4,b=3,c=2$ and $a+b+c = 4+3+2 = \boxed{9}$

Here is a solution just in case you are not able to split the expression into a perfect square.

First notice that $a<6$ and $c=2$ . Further we have $2ab=24$ or $ab=12$ .

So we must have $a=4$ and $b=3$ or $a=3$ and $b=4$ .

Finally we just have to check which one of them satisfies $a^2+b^2c=a^2+2b^2=34$

We see that $a=4$ and $b=3$ satisfies it.

Thus we have $a+b+c=4+3+2=\boxed{9}$ .

$\sqrt{(34+24\sqrt{2})}=\sqrt{(4+3\sqrt{2})^2}=(4+3\sqrt{2})$

Comparing it with $a+b\sqrt{c}$ we get, $a=4,b=3,c=2$ ,hence $a+b+c=\boxed{9}$ .