Q71 / 180 · COMEDK 2022 · Mathematics
The probability of choosing randomly a number $c$ from the set $\{1,2,3, \ldots 9\}$ such that the quadratic equation $x^2+4 x+c=0$ has real roots, is
- A$\dfrac{1}{9}$
- B$\dfrac{2}{9}$
- C$\dfrac{3}{9}$
- Correct answerD$\dfrac{4}{9}$
Solution
The quadratic equation $x^2 + 4x + c = 0$ has real roots if and only if its discriminant $D \ge 0$.
The discriminant $D$ is given by $D = b^2 - 4ac$. Here $a = 1$, $b = 4$, and $c = c$.
Substituting the values, $D = 4^2 - 4(1)(c) = 16 - 4c$.
For real roots, $16 - 4c \ge 0$, which implies $4c \le 16$, or $c \le 4$.
The set of possible values for $c$ is $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, so the total number of outcomes is $9$.
The values of $c$ from the set that satisfy $c \le 4$ are $\{1, 2, 3, 4\}$.
The number of favorable outcomes is $4$.
The probability is $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{9}$.
Answer: $\dfrac{4}{9}$
The discriminant $D$ is given by $D = b^2 - 4ac$. Here $a = 1$, $b = 4$, and $c = c$.
Substituting the values, $D = 4^2 - 4(1)(c) = 16 - 4c$.
For real roots, $16 - 4c \ge 0$, which implies $4c \le 16$, or $c \le 4$.
The set of possible values for $c$ is $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, so the total number of outcomes is $9$.
The values of $c$ from the set that satisfy $c \le 4$ are $\{1, 2, 3, 4\}$.
The number of favorable outcomes is $4$.
The probability is $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{9}$.
Answer: $\dfrac{4}{9}$
Hence, the correct answer is option: 'D'.