Square Roots(Easy)

First assume that $(\sqrt a+\sqrt b)^2= 7+2\sqrt{10}$ . Then

$\begin{aligned} a + 2\sqrt{ab} + b & = 7+2\sqrt{10}\end{aligned}$

$\implies \begin{cases} a+b = 7 & \implies b = 7-a \\ ab = 10 & \implies a(7-a) = 10 \end{cases}$

$\begin{aligned} \implies a^2 - 7a + 10 & = 0 \\ (a-2)(a-5) & = 0 & \small \blue{\text{Since }a \text{ and }b \text{ are interchangeable,}} \\ \implies a, b & = 2, 5 \end{aligned}$

Therefore, the answer is $\sqrt b + \sqrt a = \boxed{\sqrt 5 + \sqrt 2}$ .

$\begin{aligned} \sqrt{7 + 2\sqrt{10}} &= \sqrt{(5) + (2) + 2\sqrt{(5)(2)}} \\ \\ &= \sqrt{(\sqrt{5})^2 + 2\sqrt{5}\sqrt{2} + (\sqrt{2})^2} \\ \\ &= \sqrt{(\sqrt{5} + \sqrt{2})^2} \\ \\ &= \boxed{\sqrt{5} + \sqrt{2}} \end{aligned}$

It is $\sqrt { 5 } +\sqrt { 2 }$ because when you square this $\ { \sqrt { 5 } }^{ 2 }+2\sqrt { 2 } \sqrt { 5 } +{ \sqrt { 2 } }^{ 2 }=\quad 5+2\sqrt { 10 } +2=\quad 7+2\sqrt { 10 }$