Question

If m is a natural such that $m\le 5$, then the probability that the quadratic equation $x^2+mx+\frac{1}{2}+\frac{m}{2}=0$ has real roots is

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

Answer 1

The correct option is C $\frac{3}{5}$

Discriminant D of the quadratic equation
$x^2+mx+\frac{1}{2}+\frac{m}{2}=0$ is given by
$D=m^2-4(\frac{1}{2}+\frac{m}{2})$

$=m^2-2m-2$
$=(m-1{)}^2-3$
Now, as the roots of quadratic equation are real

$\Rightarrow D\ge 0\Rightarrow (m-1{)}^2\ge 3$
This is possible for $m=3,4$ and $5$.

Also, the total number of ways of choosing $m$ is $5$.
$\therefore$ The probability of the required event $=\frac{3}{5}$.