Question

The probability of choosing randomly a number \( c \) from the set \( \{1, 2, 3, \dots, 9\} \) such that the quadratic equation \( x^2 + 4x + c = 0 \) has real roots, is

• A. \( \frac{1}{9} \)
• B. \( \frac{2}{9} \)
• C. \( \frac{3}{9} \)
• D. \( \frac{4}{9} \)

Hint:

When solving for the probability of real roots, use the discriminant condition \( \Delta \geq 0 \) for quadratic equations.

Answer 1

The Correct Option is D

For the quadratic equation \( x^2 + 4x + c = 0 \) to have real roots, the discriminant must be greater than or equal to 0. The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac \] For the equation \( x^2 + 4x + c = 0 \), \( a = 1 \), \( b = 4 \), and \( c = c \), so the discriminant becomes: \[ \Delta = 4^2 - 4 \times 1 \times c = 16 - 4c \] For real roots, the discriminant must be non-negative: \[ 16 - 4c \geq 0 \] Solving this inequality: \[ 16 \geq 4c \quad \Rightarrow \quad c \leq 4 \] Thus, the possible values for \( c \) that satisfy this condition are \( c = 1, 2, 3, 4 \). These are 4 values out of the 9 possible values in the set \( \{1, 2, 3, \dots, 9\} \). Therefore, the probability is: \[ \frac{4}{9} \]