Question

A quadratic equation $a{x}^2+bx+c=0$, with distinct coefficients is formed. If $a,b,c$ are chosen from the numbers 2, 3, 5, then the probability that the equation has real roots is

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

Answer 1

Answer: (A)

Total number of ways of assigning values 2,3 , 5 to $a,b,c,=3!=6$
Now, for quadratic equation $a{x}^2+bx+c=0$ to have real roots $b^2-4ac\ge 0.$
This is possible only when $a=2,b=5,c=3$ or $a=3,b=5,c=2$
$\Rightarrow$ Required probability $-\frac{2}{6}-\frac{1}{3}$