Polynomial Roots with no complex roots Let $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_n,b_n)$ be the ordered pairs $(a,b)$ of real numbers such that the polynomial
$$p(x) = (x^2 + ax + b)^2 +a(x^2 + ax + b) - b$$has exactly one real root and no nonreal complex roots. Find $a_1 + b_1 + a_2 + b_2 + \dots + a_n + b_n.$
I have no idea how to do this. Can someone help please?

We expand the OP's polynomial and write it as
$\tag 1 p(x) = (a b + b^2 - b) + (a^2  + 2 a b) x + (a^2  + a  + 2 b) x^2  + 2 a x^3   + x^4 $
We also have (see Robert Israel's answer)
$\tag 2 (x-r)^4 = r^4 - 4 r^3 x + 6 r^2 x^2 - 4 r x^3 + x^4$
Fortunately, we can find an easy 'coefficient hook', and so $2a = -4r$, or
$$\tag 3 r =  -\frac{a}{2}$$
Plugging this back into $\text{(2)}$, we get
$$\tag 4 (x+a/2)^4 =  \frac{a^4}{16} + \frac{a^3}{2}  x + \frac{3 a^2}{2} x^2  + 2 a x^3 + x^4$$
Once again, there is an easy way to proceed, and we find
$$\tag 5 b =  \frac{a^2 - 2a}{4}$$
Once again, there is a (relatively) easy way to proceed, and we find that
$$\tag 6 \frac{a^4}{16} - \frac{a^2}{2} + \frac{a}{2} = \frac{a^4}{16}$$
must be true.
So there are at most two possible solutions:
$(a_1, b_1) = (0, 0)$
and
$(a_2, b_2) = (1, -\frac{1}{4})$
You will find that by plugging into $p(x)$ that they both work - the polynomial will only have one real root with multiplicity $4$.
I have no idea why we are interested in the sum of these coordinates. Perhaps I need to review algebra-precalculus.
