Probability that the roots of a quadratic equation are real 
Roots of the quadratic equation $x^2+5x+3=0$ are $4\sin^2\alpha+a$ and $4\cos^2\alpha+a$. Another quadratic equation is  $x^2+px+q=0$ where $p,q\in\mathbb{N}$ and $p,q\in[1,10]$. Find the probability that the roots of second  quadratic equation are real and that they are $4sin^4\alpha+b$ and $4\cos^4\alpha+b$.

$$p^2-4q\ge 0$$
If $p=1$, then no possibilities.
If $p=2$, then $q=1$.
If $p=7,8,9,10$, then $q\in[1,10]$.
But in this way there may be repetitions. I need to find the number quadratic equations first and then I can use the fact the difference in roots for both equations is same to reduce the total possibilities.
 A: As roots of the quadratic equation $x^2+5x+3=0$ are $x_1=4\cos^2\alpha+a$ and $x_2=4\sin^2\alpha+a$, so we get the magnitude of difference b/w the roots as 
$$\left|x_1-x_2\right|=\left|4\cos^2\alpha-4\sin^2\alpha\right|=\left|\pm\sqrt{5^2-4(3)}\right|=\sqrt{13}\tag{1}$$
Now consider the quadratic equation $x^2+px+q=0$, let its roots be $\delta_1$ and $\delta_2$.
For the roots to be real $D\ge0\implies p^2-4q\ge 0$. Since, $p,q\in \Bbb{N}$, so $p\ge2, q\ge 1$.
The difference of the roots is
$$\left|\delta_1-\delta_2\right|=|\pm\sqrt{p^2-4q}|=\sqrt{p^2-4q}\tag{2}$$
The roots also need be $\delta_1=4\cos^4\alpha+b$ and $\delta_2=4\sin^4\alpha+b$. So $(2)$ becomes 
$$|\delta_1-\delta_2|=\sqrt{p^2-4q}=|4\cos^4\alpha-4\sin^4\alpha|=|4\cos^2\alpha-4\sin^2\alpha|=\sqrt{13}\qquad(\text{from (1)})\\
\implies p^2-4q=13$$
And on testing for different value of $p$ and $q$, we find that only the following ordered pair satisfy the required conditions. 
$$(p,q)=\{(5,3),(7,9)\}$$
Total no. of ordered pairs $(p,q)$ are $10\times 10=100$ and only two of these satisfy the two given conditions, hence the probability is
$$P=\frac{2}{100}=\boxed{\frac{1}{50}}$$
