This means that the two quadratics have a common root. Let the roots be $r_1$ $r_2$, $r_3$. Where $r_3$ is the common root, $r_1$ is the root of the first one and $r_2$ the root of the second one. $r_1+r_3 = -a$ $r_1r_3 = b$ $r_2+r_3 = -b$ $r_2r_3 = a$ $r_2r_3 = -r_1-r_3$ $r_2r_3+r_1+r_3 = 0$ $r_1r_3 = -r_2-r_3$ $r_1r_3+r_2+r_3 = 0$ $r_2r_3+r_1+r_3 = r_1r_3+r_2+r_3$ $r_2r_3+r_1 = r_1r_3+r_2$ $r_1(1-r_3) = r_2(1-r_3)$ if $r_3 \neq 1$ then $r_1 = r_2$ which means $a = b$ so $r_1$ and $r_3$ are roots of $x^2+ax+a$ Don't know how to go from there though.

mihirb wrote: This means that the two quadratics have a common root. Let the roots be $r_1$ $r_2$, $r_3$. Where $r_3$ is the common root, $r_1$ is the root of the first one and $r_2$ the root of the second one. $r_1+r_3 = -a$ $r_1r_3 = b$ $r_2+r_3 = -b$ $r_2r_3 = a$ $r_2r_3 = -r_1-r_3$ $r_2r_3+r_1+r_3 = 0$ $r_1r_3 = -r_2-r_3$ $r_1r_3+r_2+r_3 = 0$ $r_2r_3+r_1+r_3 = r_1r_3+r_2+r_3$ $r_2r_3+r_1 = r_1r_3+r_2$ $r_1(1-r_3) = r_2(1-r_3)$ if $r_3 \neq 1$ then $r_1 = r_2$ which means $a = b$ so $r_1$ and $r_3$ are roots of $x^2+ax+a$ Don't know how to go from there though. $r_1$, $r_2$ are distinct. So $r_3 = 1$. Thus $r_1r_3+r_2+r_3 = 0$ reduces to $r_1+r_2+r_3 = 0$ and we're done.

Just saving space will post my sol later,cuz there's no proper sol here at all

$(x^2+ax+b) = (x+l)(x+k)$ $(x^2+bx+a) = (x+m)(x+n)$ $b = m+n$ $a=mn$ $a =k+l$ $b=kl$ $(x+l)(x+k)(x+m)(x+n) $ one of them must be equal to each other Case 1 $l=m$ $mn=k+l$ $m+n=kl$ $l+n=kl$ $ln=k+l$ $l=-1$,$n=1-k$ $(k-1 \ne 0)$ $m=-1$ $(-1+1-k+k) = 0$ $l= \frac{k}{k-1},n=k$(Wrong since we have three distinct real roots) (Case 2,Case 3 ,Case 4 gonna have same 2 equalities also but it is impossible since we have three distinct real roots so we only have one equalities that i mention before so im not gonna rewrite one by one again since i already give an example in case 1) Case 2 $l=n$ $n=-1$,$m=1-k$ $(-1+1-k+k) = 0$ $k-1 \ne 0$ Case 3 $k=m$ $k+l=mn$ $kl=m+n$ $ml=m+n$ $m+l=mn$ $m=-1$ ,$n=1-l$ $(l-1)\ne 0$ $(-1+1-l+l) = 0$ Case 4 $k=n$ $k+l=mn$ $kl=m+n$ $n+l=mn$ $nl=m+n$ $n=-1$,$m=1-l$ $(-1+1-l+l) = 0$ $(l-1)\ne0$ and we are done so all the possible sum of these three roots are 0

Project_Donkey_into_M4 wrote: Just saving space will post my sol later,cuz there's no proper sol here at all Well now there is a proper solution! $0$ Notice that they must have a common root, let’s say $\alpha$. Plugging $\alpha$ And subtracting the two equations we get $\alpha = 1$ Or $a=b$. $a=b$ Is rejected since the product will have $2$ Distinct solutions not 3. Now use vieta’s (lets assume that $x^2+ax+b=(x-1)(x-c)$ and $x^2+bx+a=(x-1)(x-d)$) on the first quadratic to get $a+b=-1$ And using vietas on both the quadratics we get $c=b$ And $a=d$ Which means $$c+d+\alpha=a+b+1=-1+1=0 \ \blacksquare$$