Mathematics - Quadratic Equation Question with Solution | TestHub

Let $\alpha$ be the common real root. Then ${\alpha }^2+2\alpha +a=0$ (1) and ${\alpha }^2+a\alpha +2=0$ (2). Subtracting (2) from (1): $(2-a)\alpha +(a-2)=0⟹(2-a)(\alpha -1)=0$. This yields two possibilities: $a=2$ or $\alpha =1$. Case 1: $a=2$.

Substituting $a=2$ into equation (1) gives $x^2+2x+2=0$. The discriminant is $D=2^2-4(1)(2)=-4<0$. Since $D<0$, the roots are non-real. Thus, there is no common real root if $a=2$. So $a=2$ is not a valid value. Case 2: $\alpha =1$. Substitute $\alpha =1$ into equation (1): $1^2+2(1)+a=0⟹1+2+a=0⟹a=-3$.

If $a=-3$, 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 $1,2$). Both equations have a common real root $x=1$. So $a=-3$ is a valid value.

The only possible real value of $a$ is $-3$. The sum of all possible real values of $a$ is $-3$.