If the equation $2x^2-7x+1=0$ and $a{x}^2+bx+2=0$ have a common root, then

Let $\alpha$ be a common root of the two equations. 

Then

$\begin{array}{l}2{\alpha }^2-7\alpha +1=0\\ a{\alpha }^2+b\alpha +2=0\\ \Rightarrow \frac{{\alpha }^2}{-14-b}=\frac{\alpha }{a-4}=\frac{1}{2b+7a}\\ \Rightarrow \frac{a-4}{2b+7a}=\frac{b+14}{4-a}\\ \Rightarrow (7a+2b)(b+14)+(a-4{)}^2=0\end{array}$

Clearly, $a=4,b=-14$ satisfy this equation.