Difficulties in denesting radicals $\sqrt{17+12\sqrt{2}}\,+\,\sqrt{17-12\sqrt{2}}$

Question

$\sqrt{17+12\sqrt{2}}\,+\,\sqrt{17-12\sqrt{2}}$ is the algebraic nested radicals I was given to solve without a calculator. I was told the answer is a rational number and it can be solved through algebraic manipulation.

Some progress I made so far:

$\sqrt{17+12\sqrt{2}}\,+\,\sqrt{17-12\sqrt{2}}$

= $\sqrt{\sqrt{289}+\sqrt{288}}\,+\,\sqrt{\sqrt{289}-\sqrt{288}}$

= $\sqrt{\sqrt{289}+\sqrt{289-1}}\,+\,\sqrt{\sqrt{289}-\sqrt{289-1}}$

= $\sqrt{{17}+\sqrt{17^2-1}}\,+\,\sqrt{{17}-\sqrt{17^2-1}}$

That's all I can find for now, I don't see how I can continue from here though.

Answer 1

Note that $17+12\sqrt{2}=(3+2\sqrt{2})^2$ and (therefore) $17-12\sqrt{2}=(3-2\sqrt{2})^2$.

Answer 2

I figured it out myself with the help of Andre Nicolas (Thanks!)

So $\sqrt{17+12\sqrt{2}} = \sqrt{9 + 2\times3\times2\sqrt{2} + 8}$

which is in the form of

$\sqrt{a^2+2ab+b^2}$

so it is reduced to $\sqrt{(a+b)^2}$

which easily reduces to

$a+b$

or in the question itself

$3+2\sqrt{2}$

Apply the same principles to the second part and we get

$3-2\sqrt{2}$

Adding them together we get $3+3=6$

Answer 3

I notice that one term is a sum and the other is a difference. I know that $(a + b)(a - b) = (a^2 - b^2)$ And I hope this will play out.

I have $\sqrt{\text{something}} + \sqrt{\text{something else}}$. I want to get rid of the radical signs. So the first thing I think of is to square them.

Now normally squaring a sum is a bad idea as $(a + b)^2 = a^2 + 2ab + b^2$ and the extra term $2ab$ just won't let me get anywhere. But as I noticed ones a sum and the other's a difference I hope this will somehow become a difference of squares.

$\sqrt{17+12\sqrt{2}}\,+\,\sqrt{17-12\sqrt{2}} =$

$\sqrt{ \left(\sqrt{17+12\sqrt{2}}\,+\,\sqrt{17-12\sqrt{2}} \right)^2 }=$

$\sqrt {(17+12\sqrt{2})\, + 2(\sqrt{17+12\sqrt{2}}\sqrt{17-12\sqrt{2}})+\,(17-12\sqrt{2})} =$

$\sqrt{ 34 + 2(\sqrt{17+12\sqrt{2}}\sqrt{17-12\sqrt{2}}) }=$

$\sqrt{ (34 + 2\left(\sqrt{(17+12\sqrt{2})(17-12\sqrt{2}})\right)} =$

$\sqrt{ (34 + 2(\sqrt{17^2-12^2\times2})} =$

$\sqrt{ (34 + 2(\sqrt{289-288}) }=$ (Well, that is a nice surprise!)

$\sqrt {34 + 2(\sqrt{1})} =$

$\sqrt {34 + 2} = \sqrt {36} = 6$

Answer 4

$\sqrt{17+12\sqrt{2}}\,+\,\sqrt{17-12\sqrt{2}} = a$

$a^2 = (\sqrt{17+12\sqrt{2}}\,+\,\sqrt{17-12\sqrt{2}})^2 = 36 \implies a = 6, -6$. Negative does not work as it $a = b + c$ where $c < b$.