<s>[INST] There is a single pair $(a,b)$ such that $x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$. For this pair, what is $a+b$? 
There is a single pair $(a,b)$ such that $x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$. For this pair, what is $a+b$?

I know how to find rational roots now because of my last question asked, but what about real roots? What do I do?
 [/INST] $$0=x^2+2(1+a)x+(3a^2+4ab+4b^2+2) =\\(x+1+a)^2+(3a^2+4ab+4b^2+2-1-2a-a^2)
 =\\(x+1+a)^2+(2a^2+4ab+4b^2+1-2a)
=\\(x+1+a)^2+(a^2+4ab+4b^2)+(a^2-2a+1) \\
=(x+1+a)^2+(a+2b)^2+(a-1)^2
$$
Since
$$(x+1+a)^2+(a+2b)^2+(a-1)^2=0$$
you get
$$x+1+a=0\\
a+2b=0\\
a-1=0$$
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