<s>[INST] If A and B are positive real numbers and each of the equations: $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots, what is the smallest value of A+B Problem
In the equation:$x^2+ax+2b=0$ and $x^2+2bx+a=0$, we have to figure out the sum of a and b by using the following identity:
$P(x)=Q(x)*D(x)+R(x)$
where P(x) is the equations and q(x) is the quotient and d(x) is the divisor and r(x) is the remainder.
My though process
my initial stance was that because these two equations are both equal to zero, then they must equal each other so i set 
$x^2+ax+2b=x^2+2bx+a$
which simplifies into:
$ax+2b=2bx+a$
then i knew that that didn't go anywhere so then i set x=0 in the first equation and thought that B was zero but that can't be true
So then i assigned R and S as the roots to the equations where i figured that in general given a simple quadratic equation: $ax^2+bx+c$ with the roots r and s, $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$
and going back to the real equation we can set 
$r+s=\frac{-a}{1}$ in the equation $x^2+ax+2b=0$ and $rs=\frac{2b}{1}$
and with the same thing to the other equation I got:
$r+s=\frac{-2b}{1}$ and $rs=\frac{a}{1}$
as stated these two equation can be set equal to each other because they are both equal to 0.
This means that $\frac{-a}{1}$=$\frac{-2b}{1}$
and $\frac{2b}{1}$=$\frac{a}{1}$
and thats where i got stuck.
 [/INST] HINT:
We need discriminant $\ge0,$ 
$\implies a^2\ge8b, 4b^2\ge4a\iff b^2\ge a$
$a^4\ge64a\iff a(a-4)(a^2+4a+4^2)\ge0\iff a(a-4)\ge0$ as $a^2+4a+4^2=(a+2)^2+2^2\ge2^2>0$
$\implies$ either $a\le0$ or $a-4\ge0\iff a\ge4$
But $a$ is positive.
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