The coefficients a, b, c in the quadratic equ | vedclass

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(C) The quadratic equation is $ax^2 + bx + c = 0$. Since $a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8\}$,the total number of possible triplets $(a, b, c)$ is

Vedclass · Accepted Answer

$\frac{1}{64}$

Vedclass · Answer

(C) The quadratic equation is $ax^2 + bx + c = 0$. Since $a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8\}$,the total number of possible triplets $(a, b, c)$ is $8 	imes 8 	imes 8 = 512$. For the equation to have repeated roots,the discriminant $D$ must be zero,i.e.,$b^2 - 4ac = 0$,which implies $b^2 = 4ac$. We list the possible triplets $(a, b, c)$ satisfying $b^2 = 4ac$: If $b=2$,$4 = 4ac \Rightarrow ac = 1 \Rightarrow (1, 2, 1)$. If $b=4$,$16 = 4ac \Rightarrow ac = 4 \Rightarrow (1, 4, 4), (4, 4, 1), (2, 4, 2)$. If $b=6$,$36 = 4ac \Rightarrow ac = 9 \Rightarrow (3, 6, 3)$. If $b=8$,$64 = 4ac \Rightarrow ac = 16 \Rightarrow (2, 8, 8), (8, 8, 2), (4, 8, 4)$. There are $8$ such favorable cases. The probability is $\frac{	ext{favorable cases}}{	ext{total cases}} = \frac{8}{512} = \frac{1}{64}$.

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

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