If the equation 2x2−7x+1=0 and ax2+bx+2=0 have a common root, then

Let α be a common root of the two equations. Then2α2−7α+1=0aα2+bα+2=0⇒α2−14−b=αa−4=12b+7a⇒a−42b+7a=b+144−a⇒(7a+2b)(b+14)+(a−4)2=0Clearly, a=4,b=−14 satisfy this equation.