Question

In the quadratic equation $a{x}^2+bx+c=0$ , the coefficients $a$, $b$, $c$ take distinct values from the set {1, 2, 3}. The probability that the roots of the equation are real is

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{2}{5}$

Answer 1

The correct option is B $\frac{1}{3}$

Since the roots are real, the discriminant must be non-negative:
$b^2-4ac\ge 0$
Now, $a,b,c\in \{1,2,3\}$
So, possible values of a, b, c are: $(a,b,c)=(1,3,2),(2,3,1)$
Total possible values = $3!=6$.
Therefore, probability = $\frac{2}{6}=\frac{1}{3}$