Roots of $f(x)+g(x)$ Question : Let $p,q,r,s \in \mathbb R$ such that $pr=2(q+s)$. Show that either $f(x)=x^2+px+q=0$ or $g(x)=x^2+rx+s=0$ has real roots .
My method : 
To the contrary suppose that both $f(x)$ and $g(x)$ has no real roots.
Now observe that $f(x)+g(x)=2x^2+(p+r)x+(q+s)=0$ 
Since $pr=2(q+s)$ , $f(x)+g(x)=2x^2+(p+r)x+\frac{pr}{2}=0$
$\Rightarrow f(x)+g(x)=x^2+(\frac{p}{2}+\frac{r}{2})x+\frac{p}{2}\cdot \frac{r}{2}=0$
$$\Rightarrow f(x)+g(x)=\left(x-\frac{p}{2}\right)\left(x-\frac{r}{2}\right)=0$$
So $f(x)+g(x)$ has two distinct real roots. 
Thus  $f(x)+g(x)<0$ for $x \in (\frac{p}{2},\frac{r}{2})$ if $r>p$.
But this is a contradiction as $f(x)>0$ and $g(x)>0$ for all $x \in \mathbb R $.
Thus either $f(x)=x^2+px+q=0$ or $g(x)=x^2+rx+s=0$ has real roots .
Am I correct ?
Can this be solved directly without using contradiction ?
 A: You are correct.
Another approach, still using contradiction, is note that, if none of the polynomials have real roots, then $p^2-4q<0$ and $r^2-4s<0$. Thus $0\le p^2<4q$ and $0\le r^2<4s$. Therefore $p^2r^2<16qs$, and, by hypothesis, $4(q+s)^2<16qs$, or $\frac{q+s}{2}<\sqrt{qs}$, which is a contradiction for the MG-MA (note that $0\le p^2<4q$ implies $q>0$. Also $s>0$)
A: (1). You did not state that $p\ne r$ and it is not needed. (2).Supposing that neither $f$ nor g has a real root, I do NOT observe that $f(x)+g(x)=0.$ Instead I observe that if neither $f$ nor $g$ has a real root then $\forall x\;(f(x)+g(x)>0).\;$ (3). I agree that $$\forall x\;[f(x)+g(x)=(x-p/2)(x-r/2)]$$ which implies that $$f(p/2)+g(p/2)=0$$ which implies that $$f(p/2)\leq 0 \lor g(p/2)\leq 0$$ which implies that $f(x)$ and $g(x)$ cannot both be always positive, which implies that at least one of $f,g$ has a root.(Because $\exists x\;(f(x)\leq 0\lor g(x)\leq 0).)$......... Another method is that if neither $f$ nor $g$ has a real root then $0>p^2-4 q$ and $0>r^2-4 s.$ Adding these, we get $0>p^2+r^2-4(q+s)=p^2+r^2-2 p r=(p-r)^2,$ which is impossible.  
