Question

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

Answer 1

Finding the probability that the roots of the equation $x^2+qx+(3/4)q+1=0$are real is:

Given that $q$ is an integer chosen at random in the interval $-10\le q\le 10$ .

Then number of possible outcomes in $[-10,10]=21$

Roots of $x^2+qx+(3/4)q+1=0$ are real.

$$\begin{gathered} \Rightarrow b^{2^{}}–4ac\ge 0 \\ \Rightarrow q^2–4\left(\right(3/4)q+1)\ge 0 \\ \Rightarrow q^2–3q–4\ge 0 \\ \Rightarrow (q–4)(q+1)\ge 0 \\ \Rightarrow q\ge 4,q\le -1 \end{gathered}$$

Favorable outcomes when $q\ge 4=\{4,5,6,7,8,9,10\}$

Favorable outcomes when $q\le -1=\{-1,-2,-3..\dots -10\}$

Number of favorable outcomes $=7+10=17$

Required probability $=17/21$

Hence, the probability that the roots of the equation ${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathit{q}\mathit{x}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{/}\mathbf{4}\mathbf{)}\mathit{q}\mathbf{}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}$are real is $\mathbf{17}\mathbf{/}\mathbf{21}$.