# 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?

## 2 Answers

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.

Hint: a polynomial with exactly one root must be of the form $$c (x - r)^n$$ where $$r$$ is the root and $$c$$ is some nonzero constant.