Notice that if both
are positive, then
![xy = 3 \implies √(xy) = \sqrt x \sqrt y = \sqrt3](https://img.qammunity.org/2023/formulas/mathematics/high-school/svqiaychnkhyi1icrwey6idztijp0g8dqa.png)
We also have the binomial expansion
![x + 2√(x)√(y) + y = \left(\sqrt x\right)^2 + 2 \sqrt x \sqrt y + \left(\sqrt y\right)^2 = \left(\sqrt x + \sqrt y\right)^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/s0v7v5wkxh67jwkv76yf1py064h440994f.png)
Then
![\left(\sqrt x + \sqrt y\right)^2 = (x + y) + 2√(xy) = 5 + 2\sqrt3 \\\\ \implies \sqrt x + \sqrt y = \boxed{√(5 + 2\sqrt3)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/f6qn489ud7utkerxf2455lukewe5247fyz.png)
Let's see if we can denest the radical. Suppose we could write
![√(5 + 2\sqrt3) = a + b \sqrt c](https://img.qammunity.org/2023/formulas/mathematics/high-school/a09kmieq5r98coknycon8c3kgbsbe8zpch.png)
for some non-zero integer constants
(and
). Squaring both sides gives
![5 + 2\sqrt3 = a^2 + 2ab\sqrt c + b^2c](https://img.qammunity.org/2023/formulas/mathematics/high-school/hdmk8tsqcsjs49x4fkhsd9g3u8ia7smf8n.png)
Let
and
. Then
, which gives
![a^2 + 3b^2 = 5](https://img.qammunity.org/2023/formulas/mathematics/high-school/oo50wrnlqartry75yrv19dkts5cxiqfotr.png)
![a^2 + \frac3{a^2} = 5](https://img.qammunity.org/2023/formulas/mathematics/high-school/29mk8061l2vmza2tvw1v9goijxdxiafl0u.png)
![a^4 - 5a^2 + 3 = 0](https://img.qammunity.org/2023/formulas/mathematics/high-school/n7p4fhm8tebsbp6o51d3gjs01rdhcblto5.png)
We can solve this with the quadratic formula. However, it'll lead to non-integer solutions for
, so we cannot denest after all.