oVlad wrote: Each of the quadratic polynomials $P(x), Q(x)$ and $P(x)+Q(x)$ with real coefficients has a repeated root. Is it guaranteed that those roots coincide? Yes : So we want (translating quadratics) $a(x-u)^2+b(x-v)^2=cx^2$ with $abc\ne 0$ (true quadratics) Looking at summands in $x^2$, we get $a+b=c$ Looking at summands in $x$, we get $au+bv=0$ Looking at constant summands, we get $au^2+bv^2=0$, which is $au^2+bv^2=v(au+bv)$ and so $au^2=auv$ and so $u=0$ or $u=v$ $u=0$ implies $v=0$ and the three quadratic have the same roots $u=v\ne 0$ implies $a+b=0$ and so $c=0$, discarded. Q.E.D.

very nice problem here we go: define $P(x)=\Phi(x-a)^2$ and $Q(x)=\Psi(x-b)^2$ so we get that $P(x)+Q(x)=x^2(\Phi+\Psi)-x(2a\Phi+2b\Psi)+\Phi a^2+\Psi b^2=0$ since $P(x)+Q(x)$ has coicident roots we must have $\triangle=0$ which gives $8ab\Phi \Psi=\Phi\Psi(4a^2+4b^2)$ now since we have $\Phi\cdot \Psi \neq 0$ we get $(a-b)^2=0 \implies a=b$ $\blacksquare$