The logic in radical symplification

Question

I'm having troubles while studying radicals, namely with converting expressions with the form $$\sqrt{a+b\sqrt{c}}$$ to $$a+b\sqrt{c}$$ and vice versa. When I'm dealing with these kind of problem, I usually add and subtract $\sqrt{c}^2$. In example: $$\sqrt{27+10\sqrt{2}}$$ I factor ${27+10\sqrt{2}} = 27+10\sqrt{2}+\sqrt{2}^2-2=\left(\sqrt{2}^2+5\sqrt{2}\right)+\left(5\sqrt{2}+25\right)$ ... until I get to $\sqrt{\left(\sqrt{2}+5\right)^2}$ and finally $\sqrt{2}+5$. However, either I'm missing something, or this does not always work. For instance, I cannot solve $\sqrt{113+8\sqrt{7}}$ using this technique.I would like to fully comprehend the logic behind these conversions so that I can apply it indiscriminately.

Answer 1

For a different approach, let $\alpha = \sqrt{113+8\sqrt{7}}$. Then $\alpha$ is a root of $$(x^2-113)^2-8^27=(x^2 - 2 x - 111) (x^2 + 2 x - 111)$$ whose roots are $\pm 1 \pm 4 \sqrt 7$. We just have find the correct signs. Since $\alpha$ is positive, $\alpha= \pm 1 + 4 \sqrt 7$. Now $(1 + 4 \sqrt 7)^2=113+8\sqrt{7}$ and so $\alpha= 1 + 4 \sqrt 7$.

In the general case, let $\alpha = \sqrt{a+b\sqrt{c}}$. Then $\alpha$ is a root of $x^4- 2 a x^2 +a^2 - b^2 c$. The key in this approach is that sometimes this quartic polynomial factors as $(x^2+ux-v)(x^2-ux-v)$. This happens exactly when $u^2+2v=2a$ and $v^2=a^2 - b^2 c$. We want $u,v$ integers and this places restrictions on $a,b,c$.

This approach works for $\alpha = \sqrt{27+10\sqrt{2}}$ because $\alpha$ is a root of $$(x-27)^2-10^2 2 = x^4 - 54 x^2 + 529 = (x^2 - 10 x + 23) (x^2 + 10 x + 23)$$ whose roots are $\pm 5 \pm \sqrt{2}$. As above, this yields $\alpha = 5+\sqrt{2}$.

Answer 2

Do not expect integer solutions for problems like that.

For $$\sqrt{27+10\sqrt{2}}= a+b\sqrt{2}$$

We square both sides to get $$27+10\sqrt{2}=a^2+2b^2 +2ab\sqrt 2$$

Thus we solve a system $$ 2ab=10, a^2+2b^2=27$$ and we are lucky to find $$a=5, b=1$$

Therefore $$\sqrt{27+10\sqrt{2}}= 5+\sqrt{2}$$ On the other hand for $$\sqrt{113+8\sqrt{7}} =a+b\sqrt 7$$

We need to solve $$ ab=4, a^2+7b^2=113 $$

We get $a=1$ and $b=4$ thus $$\sqrt{113+8\sqrt{7}} =1+4\sqrt 7$$