Mathematics - Quadratic Equation Question with Solution | TestHub

Let $\alpha$ be the common root. Then ${\alpha }^2+2\alpha +k=0$ (1) and ${\alpha }^2+k\alpha +2=0$ (2). Subtracting (2) from (1) gives $(2-k)\alpha +(k-2)=0⟹(2-k)(\alpha -1)=0$. This implies either $k=2$ or $\alpha =1$.

Case 1: $k=2$. The equations become $x^2+2x+2=0$ and $x^2+2x+2=0$. These are identical equations. Their discriminant is $D=2^2-4(1)(2)=-4<0$, so roots are non-real. If they have roots, they have two common roots (complex conjugates), which contradicts the condition of "exactly one common root". So $k=2$ is not a solution.

Case 2: $\alpha =1$. Substitute $\alpha =1$ into equation (1): $1^2+2(1)+k=0⟹3+k=0⟹k=-3$. For $k=-3$, the equations are $x^2+2x-3=0⟹(x+3)(x-1)=0$ (roots $1,-3$) and $x^2-3x+2=0⟹(x-1)(x-2)=0$ (roots $1,2$). These equations have exactly one common root ($x=1$). Thus, the only possible value for $k$ is $-3$. The sum of all possible values of $k$ is $-3$.