Question

Consider the quadratic equation \(x^{2}+bx+c=0\), where \(b,c \in \{1,2,3,4,5\}\). What is the probability that the equation has real roots?

• A. 0.48
• B. 0.45
• C. 0.5
• D. 0.6

Hint:

A quadratic equation has real roots whenever its discriminant \(b^2-4ac\) is greater than or equal to zero.

Answer 1

The Correct Option is A

Concept:
A quadratic equation \[ x^2 + bx + c = 0 \] has real roots if and only if its discriminant is non-negative. \[ D = b^2 - 4ac \] Since \(a = 1\), \[ D = b^2 - 4c \] For real roots, \[ b^2 - 4c \ge 0 \] \[ b^2 \ge 4c \]
Step 1: Find the total number of possible pairs \((b,c)\).
Since \[ b, c \in \{1,2,3,4,5\} \] there are \[ 5 \times 5 = 25 \] possible quadratic equations. 
Step 2: Count the favourable cases.
\[ \begin{array}{|c|c|c|} \hline b & b^2 & c \text{ satisfying } b^2 \ge 4c \\ \hline 1 & 1 & 0 \\ 2 & 4 & 1 \\ 3 & 9 & 1,2 \\ 4 & 16 & 1,2,3,4 \\ 5 & 25 & 1,2,3,4,5 \\ \hline \end{array} \] Number of favourable cases: \[ = 0 + 1 + 2 + 4 + 5 = 12 \] 
Step 3: Calculate the probability.
\[ P = \frac{12}{25} \] \[ \boxed{P = 0.48} \]