<s>[INST] Find the minimum distance from the point Find the minimum distance from the point $ \ (5,0) \ $ to the curve $ \ y=\sqrt x+2 \ $. 
Answer:
Let $ \ (x,y) \ $ be the closest point on $ \ y=\sqrt x+2 \ $ from $  (5,0) $ . 
Then the distance is given by 
$ d(x,y)=\sqrt{(x-5)^2+y^2} \ $ 
We will minimize the function $ \ g(x,y)=(x-5)^2+y^2 \ $ 
replacing $ y \ \ by \ \ \sqrt x+2 \ $ , we get 
$ g(x)=(x-5)^2+(\sqrt x+2)^2 \ $ 
The extreme points \ are 
$ f'(x)=0 \\  2(x-5)+\frac{\sqrt x+2}{\sqrt x} =0 \\ 2x \sqrt x-9 \sqrt x+2=0 $ 
This becomes complicated . I am  unable to calculate the closest point.
Help me out
 [/INST] write your last equation as $$2x-10+1+\frac{2}{\sqrt{x}}=0$$ and Isolate the term with the square root:
$$\frac{2}{\sqrt{x}}=-2x+9$$ squaring gives
$$\frac{4}{x}=(-2x+9)^2$$
can you finish?
factorizing the last equation gives $$- \left( x-4 \right)  \left( 4\,{x}^{2}-20\,x+1 \right) =0$$
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