Think about the geometry

Note that the quantity can suspiciously be written as $\sqrt{7^2 + 3^2 - 2(7)(3) x} + \sqrt{10^2 + 7^2 - 2(7)(10)\sqrt{1-x^2}}$ . The quantities under the radical remind us of the law of cosines, so let's make the substitution $x = \cos(\alpha)$ and thus $\sqrt{1-x^2} = \sin(\alpha) = \cos(\pi/2 - \alpha)$ , taking care to note that we are now optimizing over $0 \leq \alpha \leq \pi$ .

Our expression now looks like $\sqrt{7^2 + 3^2 - 2(7)(3) \cos(\alpha)} + \sqrt{10^2 + 7^2 - 2(7)(10)\cos(\pi/2 - \alpha)}$ .

Let $c = \sqrt{7^2 + 3^2 - 2(7)(3) \cos(\alpha)}$ and $c' = \sqrt{10^2 + 7^2 - 2(7)(10)\cos(\pi/2 - \alpha)}$ . Geometrically, we have the triangles $c-7-3$ and $c'-7-10$ . Note that because they have a common side and because the angle opposite to $c$ is $\alpha$ and the angle opposite to $c'$ is $\pi/2 - \alpha$ , we can put the two triangles together to obtain a quadrilateral with a right angle like in the following picture:

Clearly, $c+c'$ is minimized when $c$ and $c'$ form the hypotenuse of the right triangle with sides $3$ and $10$ , so the minimal value is $\boxed{\sqrt{109}}$ .

$\Huge \color{#D61F06}{OR}$

$\color{#3D99F6}{f(x)=\sqrt{149-140*\sqrt{1-x^2}}+\sqrt{58-42*x}. \\ f '(x)=\dfrac{70x}{(1-x^2)*\sqrt{149-140*\sqrt{1-x^2}}} -\dfrac{21}{\sqrt{58-42*x}}=0. \\ Using~~"kissan~ online~Advanced~Calculator":-\\ -1,.1,20, gave~lowest~between~.0~and~.2 .\\ 0,.01,20,gave~lowest~between~.12~and~.14 .\\ .12,.001,20,gave~lowest~between~.13~and~15 .\\ and~so~on,~I~got~upto~x=0.131163191.\\ f(0.131163191)=\sqrt{109}=\sqrt a\\ \implies~a=\Large \color{#D61F06}{109}.}$
Only the blue portion is my input. Why there is extra black portion below OR!!!!