Question

If an integer $q$ is chosen at random in interval $-10\le q\le 10,$ then the probability that the roots of the equation $x^2+qx+\frac{3q}{4}+1=0$ are real, is

• A. $\frac{17}{21}$
• B. $\frac{13}{21}$
• C. $\frac{14}{21}$
• D. $\frac{16}{21}$

Answer 1

The correct option is A $\frac{17}{21}$

Total possible selection of the integer $q$ is $21.$
Given, the roots of the equation $x^2+qx+\frac{3q}{4}+1=0$ are real.
$\therefore q^2-4(\frac{3q}{4}+1)\ge 0$
$\Rightarrow q^2-3q-4\ge 0$
$\Rightarrow (q-4)(q+1)\ge 0$
$\Rightarrow q\in (-\infty ,-1]\cup [4,\infty )\cdots \left(i\right)$
Given $q\in [-10,10]\cdots \left(ii\right)$
Using $\left(i\right)$ and $\left(ii\right),$ we get
$q\in [-10,-1]\cup [4,10]$
$\Rightarrow$ Total number of favourable values of $q=17$
$\therefore$ Required Probability $=\frac{17}{21}$