Probability of real roots(short way)

Asked Aug 03 '18 at 07:18

Active Aug 03 '18 at 12:08

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Each coefficient in the equation $ax^2+bx+c$ is determined by throwing an ordinary dice. Find the probability that the given equation will have real roots? I counted the cases which fulfill D(discriminant) greater than zero, but that turned into a long method.

Is there an elegant short way to do this question?

asked Aug 03 '18 at 07:18

user90596

How is each coefficient determined according to the die result? Also, is the die an ordinary, six sided one with numbers from $1$ to $6$ or is it arbitrarily sized? Are the coefficients $a,b,c$ real? – qualcuno Aug 03 '18 at 07:21
Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Aug 03 '18 at 07:25
Yes the die is an ordinary six sided one and as the numbers on the dice are from 1to 6 do a,b,c must be real – user90596 Aug 03 '18 at 07:27
It seems you need to consider cases $b \leq 2 \sqrt{ac}$ and $b \geq 2 \sqrt{ac}$. – Alex Aug 03 '18 at 07:28
If discriminant$ b^2-4ac > 0 $then both the condition s you gave are satisfied. – user90596 Aug 03 '18 at 07:30
@user90596: that's exactly what I wrote – Alex Aug 03 '18 at 07:31
If $D=0$ (e.g. $a=c=1, b=2$) you get a double real root – Henry Aug 03 '18 at 07:35
Oh yes , sorry I used it while solving but some how forgot all about it now, – user90596 Aug 03 '18 at 07:37
But is there a way to do it without counting cases – user90596 Aug 03 '18 at 07:38
Yeah, it does, thanks – user90596 Feb 13 '20 at 10:53

Probability of real roots(short way)

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