Question

The coefficients a, b, c in the quadratic equation ax² + 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 1

The Correct Option is C

Given the quadratic equation: \[ ax^2 + bx + c = 0 \] where \(a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8\}\).

For repeated roots, the discriminant must be zero: \[ D = 0 \implies b^2 - 4ac = 0 \implies b^2 = 4ac \]

The total number of possible choices for \((a, b, c)\) is: \[ 8 \times 8 \times 8 = 512 \]

Number of favorable cases for \(b^2 = 4ac\) is 8. Therefore, the probability is: \[ \text{Prob} = \frac{8}{512} = \frac{1}{64} \]

The possible values for \((a, b, c)\) satisfying \(b^2 = 4ac\) are: \[ (1, 2, 1), \, (2, 4, 2), \, (1, 4, 4), \, (4, 4, 1), \, (3, 6, 3), \, (2, 8, 8), \, (8, 8, 2), \, (4, 8, 4) \] This gives 8 cases.

Answer 2

Step 1: Condition for repeated roots
A quadratic equation \( ax^2 + bx + c = 0 \) has repeated (equal) roots if and only if its discriminant is zero, i.e. \[ D = b^2 - 4ac = 0. \] So we need \( b^2 = 4ac. \)

Step 2: Possible values of a, b, c
Each of \( a, b, c \) can be chosen independently from the set \(\{1, 2, 3, 4, 5, 6, 7, 8\}\).
Thus, the total number of possible ordered triples \((a, b, c)\) is \(8^3 = 512.\)

Step 3: Count favorable cases where \( b^2 = 4ac \)
We must find all integer triples satisfying \( b^2 = 4ac \) with \( a, b, c \in \{1,2,3,4,5,6,7,8\}. \)
Since \( 4ac \) must be a perfect square, let \( 4ac = k^2 \Rightarrow ac = (k/2)^2. \) Hence \( k \) must be even. Let \( k = 2m \Rightarrow ac = m^2. \) Then \( b = 2m. \) Because \( b \in \{1,\dots,8\} \), possible even values of \(b\) are 2, 4, 6, 8. So \( m = 1, 2, 3, 4.\)

Step 4: Check each possible m (and corresponding b)
Case 1: \( b = 2, m = 1 \Rightarrow ac = 1. \) Possible (a, c) = (1, 1). → 1 way.
Case 2: \( b = 4, m = 2 \Rightarrow ac = 4. \) Possible (a, c): (1, 4), (2, 2), (4, 1). → 3 ways.
Case 3: \( b = 6, m = 3 \Rightarrow ac = 9. \) Possible (a, c): (1, 9) invalid, (3, 3), (9, 1) invalid. → 1 valid way (3,3).
Case 4: \( b = 8, m = 4 \Rightarrow ac = 16. \) Possible (a, c): (2,8), (4,4), (8,2). → 3 ways.

Total favorable combinations = 1 + 3 + 1 + 3 = 8.

Step 5: Compute probability
\[ P = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{8}{512} = \frac{1}{64}. \]

Final answer

\(\frac{1}{64}\)