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Let $ax^2 + bx+c =0$

A fair six-faced die is thrown three times to determine the values of a, b and c.

What is the probability of the root being real?

My solution:

The probability is number of triples (a,b,c) such that $b^2-4ac \ge0$ over $6^3$

I ran a computer program and found out that there are 43 triples that have this property. Hence the probability is $\frac{43}{6^3}=0.1991$.

My question is, how could I have found all values (a, b, c) without the use of brute force?

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Travis
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  • Do you mean that there are 43 triples that have the property? – gary Jul 28 '17 at 18:19
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    A non-brute force method would look something like: Let $d=\lfloor\frac{b^2}4\rfloor$ then number of triples is $\min(b-1,1) + 2\cdot\max(\min(d-1,5),0)+\max(\min(d-2,1),0)+\max(d-4,0)$. Much easier to solve it by brute force. – Χpẘ Jul 28 '17 at 22:21

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I would divide into cases on $b$. The factor $4ac$ wants to be too large. For $b=1$ there are no possibilities. For $b=2$ we can only have $a,c=11$. For $b=3$ we can only have $11,12,21$. For $b=4$ we can only have $11,12,13,14,21,22,31,41$ and so on. It isn't too much work if you are systematic. It is still "brute force"

Ross Millikan
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