# Probability that the equation so formed will have real roots

Three dice named A,B,C are thrown and the numbers shown on them are put for $a,b,c$ respectively in the quadratic equation $ax^2+bx+c=0$.What is the probability that the equation so formed will have real roots?

Since the roots are real.Therefore,$b^2\geq 4ac$

When $b=2$,then $a=1,c=1$;When $b=3$,then $a=2,c=1$;When $b=3$,then $a=1,c=1$
When $b=3$,then $a=1,c=2$;When $b=4$,then $a=2,c=2$;When $b=4$,then $a=1,c=4$
When $b=4$,then $a=4,c=1$;When $b=4$,then $a=1,c=3$;When $b=4$,then $a=3,c=1$
When $b=4$,then $a=1,c=2$;When $b=4$,then $a=2,c=1$;When $b=4$,then $a=1,c=1$
When $b=5$,then $a=2,c=3$;When $b=5$,then $a=3,c=2$;When $b=5$,then $a=5,c=1$
When $b=5$,then $a=1,c=5$;When $b=5$,then $a=2,c=2$;When $b=5$,then $a=3,c=1$
When $b=5$,then $a=1,c=3$;When $b=5$,then $a=2,c=1$;When $b=5$,then $a=1,c=2$
When $b=5$,then $a=1,c=1$;When $b=6$,then $a=3,c=3$;When $b=6$,then $a=2,c=4$
When $b=6$,then $a=4,c=2$;When $b=6$,then $a=2,c=3$;When $b=6$,then $a=3,c=2$
When $b=6$,then $a=5,c=1$;When $b=6$,then $a=1,c=5$;When $b=6$,then $a=2,c=2$
When $b=6$,then $a=3,c=1$;When $b=6$,then $a=1,c=3$;When $b=6$,then $a=2,c=1$
When $b=6$,then $a=1,c=2$;When $b=6$,then $a=1,c=1$

I have counted 35 cases and my probability is $\frac{35}{216}$,but in the book $\frac{43}{216}$ is the answer.

Which 8 cases i have left,i checked but could not find.Please help me.

#### Solutions Collecting From Web of "Probability that the equation so formed will have real roots"

You seem to be missing the cases when $b= 5, 6$ and $a, c = 1$ and $c, a = 4, 6$.

It helps to have a partially-filled in multiplication table handy:

\begin{array}{c|cccccc}
& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
1 & 1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 4 & 6 & 8 & & \\
3 & 3 & 6 & 9 & & & \\
4 & 4 & 8 & & & & \\
5 & 5 & & & & & \\
6 & 6 & & & & & \\
\end{array}

Then we can simply read off the pairs of numbers $(a,c)$ that satisfy
$ac \leq \frac14 b^2$ for each value of $b$:

\begin{align}
ac \leq \left\lfloor\tfrac14(2^2)\right\rfloor = 1:\qquad & (1,1) \\
ac \leq \left\lfloor\tfrac14(3^2)\right\rfloor = 2:\qquad & (1,1),(1,2),(2,1) \\
ac \leq \left\lfloor\tfrac14(4^2)\right\rfloor = 4:\qquad &
(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(3,1),(4,1) \\
ac \leq \left\lfloor\tfrac14(5^2)\right\rfloor = 6:\qquad &
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1),(2,2),(2,3), \\
& (3,1),(3,2), (4,1),(5,1),(6,1) \\
ac \leq \left\lfloor\tfrac14(6^2)\right\rfloor = 9:\qquad &
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1),(2,2),(2,3),(2,4) \\
& (3,1),(3,2),(3,3), (4,1),(4,2), (5,1),(6,1) \\
\end{align}

As you can see, there are $14$ pairs for $b=5$ and $17$ pairs for $b=6$,
but you have listed only $10$ rolls with $b=5$ and only $13$ with $b=6$.
That accounts for your eight missing rolls.
If you compare the lists you can discover which rolls you missed.

In the following $5$ cases equality holds:

$$\begin{matrix} A B C\\ 1 2 1\\ 1 4 4\\ 2 4 2\\ 3 6 3\\ 4 4 1\\ \end{matrix}$$

In $43$ cases the inequality holds. $43-5=38$, so you missed $3$ cases.

Ooops: I can see that Paul Sinclair has found the missing cases.