Simplification of radicals: $\sqrt{4+\sqrt5}-\sqrt{\frac{4+\sqrt{11}}{2}}$

Question

I have a simplification question. It may be simple, but I can't solve it. Can someone give an idea ? $$\sqrt{4+\sqrt5}-\sqrt{\frac{4+\sqrt{11}}{2}}$$ I tried to rewrite it $$\sqrt{4+\frac22\sqrt5}-\sqrt{\frac{4+\sqrt{11}}{2}}$$ but can't go further. I am not looking for a complete solution, but just a clue.

Answer 1

Let $\sqrt{4+\sqrt{5}}=\sqrt{(a+b)^2}=\sqrt{a^2+b^2+2ab}$ with $a^2+b^2=4$ and $2ab=\sqrt{5}$.

$${\begin{cases}a^2+b^2=4\\2ab=\sqrt{5}\end{cases}\Rightarrow {\begin{cases}a^2+b^2=4\\4a^2b^2=5\end{cases}}\Rightarrow {\begin{cases}a^2+b^2=4\\a^2b^2=1.25\end{cases}}}\Rightarrow a^2(4-a^2)=1.25$$

Choose just one correct solution for $a$ and $b$, you can now symbolize the expression.

Answer 2

$1$st intuitive step is to see how squared expression will look like (to kill part of square roots): $$ \left(\sqrt{4+\sqrt{5}} - \sqrt{\dfrac{4+\sqrt{11}}{2}}\right)^2 \\ = 4+\sqrt{5} + 2 + \dfrac{\sqrt{11}}{2} - \sqrt{32+8\sqrt{5}+8\sqrt{11}+2\sqrt{55}}; $$

$2$nd intuitive step is try to find full square under the square root: $$ 32+8\sqrt{5}+8\sqrt{11}+2\sqrt{55}=\left(a+b\sqrt{5}+c\sqrt{11}\right)^2 $$

Further steps are for you (small hint: $a,b,c$ are positive integer numbers).

Answer 3

Try to find $\sqrt{4+\sqrt{5}}$: Suppose $(a+b\sqrt{5})^2 = \sqrt{4+\sqrt{5}}.$ Then

$$a^2+5b^2 +2ab\sqrt{5} = 4 +\sqrt{5}.$$

So $2ab = 1$ and $a^2+5b^2 = 4$. Then $b=\frac{1}{2a}$ and we have

$$a^2 + 5\left(\frac{1}{2a}\right)^2 = 4$$

or

$$a^4 -4a^2+\frac{5}{4}$$

which gives us $$a^2 = \frac{4\pm \sqrt{11}}{2}.$$

I choose the '+'. Then

$$b^2 = \frac{1}{4a^2} = \frac{1}{2(4+\sqrt{11})} = \frac{4-\sqrt{11}}{10}.$$

So then

$$\sqrt{4+\sqrt{5}} - \sqrt{\frac{4+\sqrt{11}}{2}} = a+b\sqrt{5} - a = \sqrt{\frac{4-\sqrt{11}}{10}}\sqrt{5} = \sqrt{\frac{4-\sqrt{11}}{2}} .$$

Answer 4

Use the identity that $\sqrt{a\pm b\sqrt{c}}=\sqrt{(a+d)/2}\pm \sqrt{(a-d)/2}$, where $d=\sqrt{a^2-b^2c}$. So using that here, $$\sqrt{4+\sqrt5}-\sqrt{\frac{4+\sqrt{11}}{2}}$$ $$=\left(\sqrt{\frac{4+\sqrt{4^2-5}}{2}}+\sqrt{\frac{4-\sqrt{4^2-5}}{2}}\right)-\sqrt{\frac{4+\sqrt{11}}{2}}$$ $$=\left(\sqrt{\frac{4+\sqrt{11}}{2}}+\sqrt{\frac{4-\sqrt{11}}{2}}\right)-\sqrt{\frac{4+\sqrt{11}}{2}}$$ $$=\sqrt{\frac{4-\sqrt{11}}{2}}$$