The coefficients a, b, and c in the quadratic equation ax^2 + bx… — Solution | JEE Mathematics

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The coefficients $a$ , $b$ , and $c$ 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:

A
$\frac{3}{256}$
B
$\frac{1}{128}$
C
$\frac{1}{64}$
D
$\frac{3}{128}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The coefficients $a$ , $b$ , and $c$ are chosen from $\{1,2,3,4,5,6,7,8\}$ for the quadratic equation $ax^2 + bx + c = 0$ .

Find: The probability that the equation has repeated roots.

For repeated roots, the discriminant must be zero:

b^2 - 4ac = 0

The total number of possible choices of $a$ , $b$ , and $c$ is:

8 \times 8 \times 8 = 512

From the solution, the number of favorable cases satisfying $b^2 = 4ac$ is $8$ .

Therefore, the required probability is:

\frac{8}{512} = \frac{1}{64}

Hence, the correct option is C.

Common mistakes

Using the condition for distinct roots instead of repeated roots. For repeated roots, the discriminant must satisfy $b^2 - 4ac = 0$ , not merely be positive or negative. Always start with the repeated-root criterion.
Counting the total cases incorrectly as combinations instead of ordered selections. Since $a$ , $b$ , and $c$ are chosen independently, the total number of cases is $8^3 = 512$ . Do not use unordered counting here.
Ignoring that different ordered triples $(a,b,c)$ are different outcomes. Even if two triples contain the same numbers in a different order, they are counted separately because they give different coefficients.

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Answer

Step-by-step solution

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