Mathematics - Quadratic Equation Question with Solution | TestHub

For the quadratic equation $A{x}^2+Bx+C=0$ to have real and distinct roots, its discriminant $D=B^2-4AC$ must be greater than 0. Here, $A=1$, $B=-2(k+1)$, and $C=k^2+4$. So, $D=[-2(k+1){]}^2-4(1)(k^2+4)>0$. This simplifies to $4(k^2+2k+1)-4(k^2+4)>0$, which means $k^2+2k+1-k^2-4>0$. Therefore, $2k-3>0$, which implies $2k>3$, or $k>3$. The set of all possible values of $k$ is $(3,\infty )$.