Question

If p and q are chosen randomly from the set $\{1,2,3,4,5,6,7,8,9,10\}$ with replacement. The probability that the roots of $x^2+px+q=0$ are

Answer 1

The roots f $x^2+px+q=0$ will be imaginary if and only if $p^2-4q<0$, i.e., $p^2<4q$
We enumerate the possible values of p and q for which this can happen in Table

q	p	Number of pairs of p, q
1	1	1
2	1, 2	2
3	1, 2, 3	3
4	1, 2, 3	3
5	1, 2, 3, 4	4
6	1, 2, 3, 4	4
7	1, 2, 3, 4, 5	5
8	1, 2, 3, 4, 5	5
9	1, 2, 3, 4, 5	5
10	1, 2, 3, 4, 5, 6	6
Total		38

Thus, the number of possible pairs $=38$. Also, the total number of possible pairs is $10\times 10=100$.
A) Probability for imaginary roots $P=0.38$
B)The roots of $x^2+px+q=0$ will be equal if and only if $p^2-4q=0$.
The possible pairs of (p, q) are (2, 1), (4, 4), (6, 9).
Thus, probability for equal roots $P=0.03$.
C) Probability for real roots $P=0.62$
D) As p > 0, q > 0, roots of
$x^2+px+q=0$ are negative if these are real.
i.e probability is $P=0.62$