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:

$(a_1, b_1) = (0, 0)$


$(a_2, b_2) = (1, -\frac{1}{4})$

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.