# 1995 AIME Problems/Problem 5

## Problem

For certain real values of $a, b, c,$ and $d_{},$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots. The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$ Find $b.$

## Solution 1

Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be $m,n$. Since $m+n$ is not real, $m,n$ are not conjugates, so the other pair of roots must be the conjugates of $m,n$. Let $m'$ be the conjugate of $m$, and $n'$ be the conjugate of $n$. Then, $$m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.$$ By Vieta's formulas, we have that $b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = \boxed{051}$.

## Solution 2

Let's assume that the 2 roots multiplied together are p+qi, and r+si, and the two roots added together are the conjugates of the previous roots. Using Vieta, we get $$b = (p+qi)(r+si) + (p+qi)(r-si) + (p-qi)(r+si) + (p-qi)(r-si) + (p+qi)(p-qi) + (r+si)(r-si) =$$

$$(p+qi+p-qi)(r+si+r-si) + (p+qi)(p-qi) + (r+si)(r-si) =$$

$$(2p)(2r) + p^2 + q^2 + r^2 + s^2 = 4pr + p^2 + q^2 + r^2 + s^2 = (p+r)^2 + 2pr + q^2 + s^2$$

We are now stuck. We can't simplify further. But, we look back to the problem and see the the two roots that are multiplied together give a product of $13+i$, and the two roots that are added give $3+4i$. This gets three equations nessacary for solving the problem. $$p+r = 3$$ $$pr-qs = 13$$ $$-q-s = 4$$ So, alright. Let's use the first equation to get that $(p+r)^2 = 9$, and substitute that in. Now, the equation becomes:

$$b = 9 + 2pr + q^2 + s^2$$

We wish that we can turn the 2pr into 2qs. Fortunatly, we can do that. By using the second equation, we can manipulate it to be $pr = 13+qs$, and substitute that in.

$$b = 9 + 2(13+qs) + q^2 + s^2 = 9 + 26 + 2qs + q^2 + s^2 = 35 + (q+s)^2$$

We can square both sides of the third equation, and get $(q+s)^2 = 16$ We substitute that in and we get

$$b = 35+16 = \boxed{051}$$

- AlexLikeMath