My first problem in 2016!

$f(x)=a(x-{ x }_{ 1 })(x-{ x }_{ 2 })$ . Therefore, $x_{ 1 }$ and $x_{ 2 }$ are the roots of f(x). It is given that $g(x_{ 1 })=0$ . Therefore, $dx_{ 1 }+c=0\quad \Leftrightarrow \quad c=-dx_{ 1 }$ . F(x)=f(x)+g(x), therefore F( $x_{ 1 }$ )=0. We are told F(x) has only one root, and now we know that root to be $x_{ 1 }$ . $F(x)=a(x-x_{ 1 })(x-{ x }_{ 2 })+dx+c$ . It is clear even without expanding that the coefficient of the x squared term in F(x) will be a. Therefore, knowing the roots of F(x) and the coefficient of it's x squared term, we can say, $F(x)=a(x-{ x }_{ 1 })^{ 2 }$ . $F(x_{ 2 })=f(x_{ 2 })+g(x_{ 2 })$ . We know that $x_{ 2 }$ is a zero of f(x), so $F(x_{ 2 })=g(x_{ 2 })$ . $F(x_{ 2 })=a(x_{ 2 }-x_{ 1 })^{ 2 }. g(x_{ 2 })=dx_{ 2 }+c=d{ x }_{ 2 }-dx_{ 1 }=d(x_{ 2 }-x_{ 1 })$ . Therefore, $d(x_{ 2 }-x_{ 1 })=a(x_{ 2 }-x_{ 1 })^{ 2 }\\ d=a(x_{ 2 }-x_{ 1 })$ .