The equation [tex]x^2+kx=5x+3[/tex] must have only one solution for the line [tex]y[/tex] to be a tangent line to the quadratic function [tex]f(x)[/tex].
[tex]x^2+kx=5x+3\\
x^2+kx-5x-3=0\\
x^2+(k-5)x-3=0\\
\Delta=(k-5)^2-4\cdot1\cdot(-3)=(k-5)^2+12\\\\
(k-5)^2+12=0\\
(k-5)^2=-12\\
k\in \emptyset
[/tex]
So there isn't such value of k.
