Answer
$$a = \frac{4}{{25}},{\text{ }}\sqrt {\frac{5}{{29}}} \approx 0.041552$$
Work Step by Step
$$\eqalign{
& \sqrt {\frac{5}{{29}}} \cr
& {\text{Rewrite }}\frac{5}{{29}}{\text{ as }}\frac{4}{{25}} + \frac{9}{{725}} \cr
& {\text{Using the function }}f\left( x \right) = \sqrt x ,{\text{ }} \cr
& f\left( x \right) = \sqrt x \cr
& {\text{we have that }}\frac{5}{{29}} = \frac{4}{{25}} + \frac{9}{{725}},{\text{ let }}a = \frac{4}{{25}} \cr
& {\text{* Evaluate }}f\left( {\frac{4}{{25}}} \right) \cr
& f\left( {\frac{4}{{25}}} \right) = \sqrt {\frac{4}{{25}}} \cr
& f\left( {\frac{4}{{25}}} \right) = \frac{2}{5} \cr
& \cr
& {\text{Differentiating we obtain}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt x } \right] \cr
& f'\left( x \right) = \frac{1}{{2\sqrt x }} \cr
& {\text{* Evaluate }}f'\left( {\frac{4}{{25}}} \right) \cr
& f'\left( {\frac{4}{{25}}} \right) = \frac{1}{{2\sqrt {4/25} }} \cr
& f'\left( {\frac{4}{{25}}} \right) = \frac{5}{4} \cr
& \cr
& {\text{Using the definition of a linear approximation to }}f{\text{ at }}a{\text{ }}\left( {page{\text{ 282}}} \right) \cr
& L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right),{\text{ for an interval }}I{\text{ containing }}a \cr
& \cr
& {\text{The linear approximation }}L\left( x \right){\text{ at }}a{\text{ is:}} \cr
& L\left( x \right) = \frac{2}{5} + \frac{5}{4}\left( {x - \frac{4}{{25}}} \right) \cr
& L\left( x \right) = \frac{2}{5} + \frac{5}{4}x - \frac{1}{5} \cr
& L\left( x \right) = \frac{5}{4}x + \frac{1}{5} \cr
& \cr
& {\text{Estimating the given value at }}f\left( {\frac{5}{{29}}} \right) \cr
& L\left( x \right) = \frac{5}{4}x + \frac{1}{5} \cr
& L\left( {\frac{5}{{29}}} \right) = \frac{5}{4}\left( {\frac{5}{{29}}} \right) + \frac{1}{5} \cr
& L\left( {\frac{5}{{29}}} \right) = \frac{{241}}{{580}} \cr
& {\text{Therefore}}{\text{,}} \cr
& \sqrt {\frac{5}{{29}}} \approx L\left( {\frac{5}{{29}}} \right) \approx \frac{{241}}{{580}} \approx 0.41552 \cr} $$
