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Could any one help me to solve this problem?

Given that the coefficients of the equation $ax^2+bx+c=0$ are selected by throwing an unbiased die, we need to find what is the probability of the equation having equal roots.

Thank you for Hints.

Martin Sleziak
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Myshkin
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2 Answers2

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Hint: The roots are the same iff $b^2-4ac=0$ so find the number of triples $(a,b,c)$ with each of $a,b,c$ in $\{1,2,3,4,5,6\}$ for which this equation holds, then divide by $6^3$.

OP has asked about finding the $a,b,c$: First from $b^2=4ac$ we see that $b$ must be even, so that it is one of $2,4,6$

$b=2$ (so that $b^2=4=4\cdot 1$) implies $ac=1$ giving the single pair $(1,1)$ for $(a,c).$

$b=4$ (so that $b^2=16=4\cdot 4$) implies $ac=4$ giving the three pairs $(1,4),(2,2),(4,1)$ for $(a,c)$

Finally $b=6$ (so that $b^2=36=4\cdot 9$ implies $ac=9$ giving one pair $(3,3)$ for $(a,b)$. [we can't have $(1,9),(9,1)$ since dice only go to $6$]

So the list of triples $(a,b,c)$ for which $b^2=4ac$ is $$(1,2,1),(1,4,4),(2,4,2),(4,4,1),(3,6,3).$$

coffeemath
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  • well I appreciate your hint, could you tell me what is the easiest way to find out those triplet – Myshkin Jul 03 '13 at 14:20
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for this first you have to find the maximum number of possibility i.e 6*6*6 = 6^3

for having equal roots: b^2 = 4ac

so it means that the value of 4ac must be equal to {1, 4, 9, 16, 25, 36}

so the possibilities of a and c to take the values are 1 for (1,1) & 1 for (2,2) &1 for (3,3) and for different 1 for (1,4), 1 for (4,1) so in sum gives = 5 possibilities

so 5/6^3 according to me is the answer...

Zeeshan
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