Question

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 $b.$

Answer

Answer: 51

Concepts:complex number Vieta’s Formulas polynomial

Difficulty rating: 2110

Small Hint:

Because the polynomial has real coefficients, its non-real roots occur in conjugate pairs

Big Hint:

Group the six pairwise products into the products within the two groups and the four cross-products

Solution:

Let the first two roots be $\alpha$ and $\beta.$ Since $\alpha\beta=13+i$ is not real, they are not conjugates, so the other roots are $\overline\alpha$ and $\overline\beta.$ Hence $\alpha+\beta=3-4i$ and $\overline\alpha\,\overline\beta=13-i.$ By Vieta’s formulas, $\begin{aligned}b&=\alpha\beta+\overline\alpha\,\overline\beta\\&\quad+(3-4i)(3+4i)\\&=26+25=51.\end{aligned}$

1995 AIME Problem 5

Answer: 51

Difficulty rating: 2110

Small Hint:

Big Hint:

Solution:

Problem 5 in Other Years