Animated Solution for Mathematics - Probability: The coefficients $a,b,c$ in the quadratic equation $a{x}^2+bx+c=0$ are chosen from the set $\{1,2,3,4,5,6,7,8\}$. The probability of this equation having repeated roots is :

Imagine you are standing before a quadratic equation, $a{x}^2+bx+c=0$. You have a bag of numbers, the set $\{1,2,3,4,5,6,7,8\}$, and you are tasked with picking three coefficients to build this equation.

We want to determine the probability that this equation has repeated roots. This is a puzzle of constraints and possibilities.

Every quadratic equation carries a secret identity, hidden within its discriminant, $D=b^2-4ac$. When we demand that the roots be repeated, we are essentially demanding that the parabola touches the x-axis at exactly one point.

Mathematically, this forces the discriminant to vanish: $D=0$. This gives us our guiding light:

This is the constraint that will filter our universe of possibilities.

Before we hunt for the 'special' equations, we must understand the 'total' world. We are choosing three coefficients, $a$, $b$, and $c$, each from a set of $8$ numbers.

Since each choice is independent, the total number of possible equations is:

This is our sample space. It is a large number, but we are about to shrink it down significantly.

Look closely at our condition: $b^2=4ac$. The right side is a multiple of $4$.

This is a massive clue! It tells us that $b^2$ must also be a multiple of $4$. If you test the numbers, you will realize that only even numbers, when squared, produce a multiple of $4$.

Thus, $b$ must be an even number from our set: $b\in \{2,4,6,8\}$. This realization is the key that unlocks the door.

Let us walk through the cases for $b$:

If $b=2$, then $b^2=4$. Our condition becomes $4=4ac$, or $ac=1$. The only way to get a product of $1$ with our set is $(1,1)$. This gives us one triplet: $(1,2,1)$.

If $b=4$, then $b^2=16$. Our condition becomes $16=4ac$, or $ac=4$. We need pairs $(a,c)$ that multiply to $4$, which are $(1,4)$, $(4,1)$, and $(2,2)$. This gives us three triplets: $(1,4,4)$, $(4,4,1)$, and $(2,4,2)$.

If $b=6$, then $b^2=36$, so $ac=9$. The only pair is $(3,3)$. This gives us one triplet: $(3,6,3)$.

If $b=8$, then $b^2=64$, so $ac=16$. The pairs are $(2,8)$, $(8,2)$, and $(4,4)$. This yields three more triplets: $(2,8,8)$, $(8,8,2)$, and $(4,8,4)$.

We sum our favorable outcomes: $1$ (from $b=2$) $+3$ (from $b=4$) $+1$ (from $b=6$) $+3$ (from $b=8$). Thus, the number of favorable outcomes is $n(E)=8$.

The probability is the ratio of favorable outcomes to the total sample space:

Simplifying this fraction, we arrive at the final result: