I am working through some of the quizes on brilliant.org
I came across this question.
Suppose that the following cubic polynomial has one rational root and two non-real complex roots:
$$ x^3 – 13x^2 + (40 + a)x – 5a $$
where $a$ is a positive prime number. What is the minimum value of $a$?
So, what I did was look at the discriminant, which is the long expression $18abcd -4b^3d + b^2c^2 – 4ac^3 – 27a^2d^2$. This simplifies to $-4a^3 + 184a^2 -2820a +14400.$
Then we need to determine when this is under zero, because that gives one real root.
So I need to factor this beast. I would need to get very lucky to guess at factors of this. Even using the factorization $4(14400) = 2^8 3^2 5^2 $, it is mostly guess work. So, it is probably best just to use numerical software or the solution formula directly.
Either way, we can arrive at $$ -4(a – 16)(a – 15)^2 < 0.$$ This only happens when $a > 16$, so we use the next available prime.
Finally, I arrive at the answer of $a = 17$.
The Problem
I am sure that this was meant to be done without numerical software. I am thinking there is some creative method that will give the same result. Can someone provide a hint?
Best Answer
Hint:
Observe that $x=5$ is a root of this equation. So Now you can factor $$x^3 - 13x^2 + (40 + a)x - 5a=(x-5)(x^2-8x+a)$$ Now you can look for the discriminant of the quadratic (which will be easier). This will have real roots iff $64-4a \geq 0$.