Probability of winning - and some algebra

Question

Here's a little probability riddle one of my friends proposed a weeks ago-

In a game, Sheldon is going to pick three non zero real numbers and Leonard is going to arrange them as the coefficients of a quadratic equation, ax²+bx+c=0. Sheldon wins the game only if the resulting equation has two distinct rational solutions - else Leonard wins.

What is the maximum probability of Sheldon winning the game?

This question has been bothering me for a while, and I've not reached any conclusions.

I tried to check the probabilities of winning by selecting random triplets {a,b,c} such as {1,2,3}, {3,4,5} etc but couldn't find any pattern.

Could someone please help me with the problem? A detailed solution, or a method to solve?

P.S. Additional question from the comments section: I think the problem is more interesting if you remove the probability. Is there any triple of non-zero reals such that all 6 possible quadratics have distinct rational roots?

EDIT-

Correctly pointed out by @lulu, the question should be rephrased as - "Suppose Sheldon chooses a triple of non-zero real numbers (not probabilistically). Leonard then chooses one of the 6 associated quadratics uniformly at random. What is the maximal probability for a Sheldon win?"

Answer 1

It's 0%, Leonard cant win if Sheldon chooses right numbers.

To have two distinct rational solutions delta must be greater than 0, which means b2 - 4ac > 0 => b2 > 4ac If Sheldon chooses 10 -1 -2 , he wins. b2 is always greater than 0, so a*c can't be negative, therefore a and b have to be -1 and -2. 100 > 8, Sheldon wins.

Answer 2

The way to do this is to verify the determinant of the second degree equation. You have that $ax^2+bx+c=0$ will have two distinct solution if and only if $\Delta=\sqrt{b^2-4ac}>0$. Hence what you are looking for is when $b^2-4ac>0$.

Answer 3

Consider the triple $(-3,2,1)$. Then each of the $6$ polynomials that can be formed with these coefficients has two distinct rational roots. Thus Sheldon can assure a victory by choosing this triple.

This is easily checked by hand. To speed the calculation, note that $ax^2+bx+c$ has two distinct rational roots iff $cx^2+bx+a$ does (trusting that none of the coefficients are $0$). Thus there are really only three polynomials to check.