Let a,b and c denote the outcome of three independent rolls of a fair tetrahedral die, who…

Question

Let $a,b$ and $c$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked $1,2,3,4$. If the probability that $a{x}^{2}+bx+c=0$ has all real roots is $\frac{m}{n},gcd(m,n)=1$, then $m+n$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 19

Accepted Answer

A quadratic equation $a{x}^{2}+bx+c=0$ has real roots if and only if its discriminant is non-negative. The discriminant $\Delta$ of the quadratic equation is given by:

$\Delta={b}^{2}-4ac$

For the quadratic equation to have all real roots, the discriminant must be non-negative:

$\Delta\ge 0$

That means:

${b}^{2}-4ac\ge 0$

Given that $a,b,c$ are the outcomes of rolling a fair tetrahedral die, they can each be one of the numbers 1, 2, 3, or 4. Our task is to determine the probability that this condition holds.

We need to analyze the cases where ${b}^{2}\ge 4ac$.

Let's consider all possible values for $a$, $b$, and $c$, and count how many of them satisfy the condition. Since there are 4 choices for each of the variables, there are a total of $4	imes 4	imes 4=64$ possible combinations.

Now, we count the valid combinations where ${b}^{2}\ge 4ac$:

For $a=1$: ${b}^{2}\ge 4c$

$b=1:1\ge 4c	o c\ge 1$
 $b=2:4\ge 4c	o c\le 1	o c=1$ (1 case)
 $b=3:9\ge 4c	o c\le 2	o c=12$ (2 cases)
 $b=4:16\ge 4c	o c\le 4	o c=1,2,3,4$ (4 cases)

Total for $a=1=1+2+4=7$

For $a=2$: ${b}^{2}\ge 8c$

$b=1:1\ge 8c	o$
 $b=2:4\ge 8c	o$
 $b=3:9\ge 8c	o c\le 1$ (1 case)
 $b=4:16\ge 8c	o c\le 2$ (2 cases)

Total for $a=2=1+2=3$

For $a=3$: ${b}^{2}\ge 12c$

$b=1:1\ge 12c	o$
 $b=2:4\ge 12c	o$
 $b=3:9\ge 12c	o$
 $b=4:16\ge 12c	o c\le 1$ (1 case)

Total for $a=3=1$

For $a=4$: ${b}^{2}\ge 16c$

$b=1:1\ge 16c	o$
 $b=2:4\ge 16c	o$
 $b=3:9\ge 16c	o$
 $b=4:16\ge 16c	o c\le 1$ (1 case)

Total for $a=4=1$

Adding up all the valid cases:

$7+3+1+1=12$

The total number of valid combinations is 12 out of 64. Thus, the probability is:

$\frac{12}{64}=\frac{3}{16}$

The value of $m=3$ and $n=16$. The sum $m+n=3+16=19$.

Hence, the answer is 19.

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