Q: Minimum no of balls in the box. 
A box contain more than $100$ balls of  which some are red and remaining are blue. If we take two balls at random ,the probability of getting one blue and one red is $\frac{1}{2}$. The minimum no of balls in the box?

I am not getting the way to relate the two givens in the problem. Any hints?

A: Let the no. Of red balls be x and total no. Of balls be t
So probability of selecting one red ball one blue ball is $2x(t-x)/t.(t-1)$ =$1/2$
It will form a  quadratic in t
$t^2$-(1+4x)t+4$x^2$ =0
t = (1+4x + (1+8x)^1/2)/2
For some x value u will get 1+8x in every odd no. Square so u require t around 100 , so at x= 45 u get t = 100
If u need more than 100 u can check 1+8x=$21^2$ x=55
At x=55 , t=121
I thought min 100 I am mistaken.

A: Suppose we have $r$ red balls and $b$ blue balls and $t$ be the total number of balls. We are given $t = r+b>100$. The probability of getting one blue and one red ball is $2\frac{r}{r+b}\frac{b}{r+b-1} = 2\frac{r(t-r)}{t(t-1)}$. This quantity should be $\frac{1}{2}$. Re arranging we obtain ( or at least I obtained ):
$$t^2 - t(1+4r)+4r^2 = 0$$
We want the first integer solution to this equation with $t>100$. Solving for $t$, we get:
$$2t = (1+4r) \pm \sqrt{1+8r}$$
Since the $LHS$ is even and so is $1+4r$, $\sqrt{1+8r}$ needs to be an odd square number say $m^2$. Then we are looking at integer solutions of $1+8r = m^2$. This means that:
$$m^2-1 = (m+1)(m-1) = 8r$$, since $m$ is odd, $m+1$ and $m-1$ are consecutive even numbers, so one of them is always divisible by $4$ and $m^2-1$ is always divisible by $8$. We can write $m = 2n+1$ and we obtain:
$$(n+1)n = 2r$$
So we can choose $2r$, to be the product of any two consecutive numbers. Writing the $LHS$ of the equation for $2t$ in terms of $n$ we obtain:
$$2t = 1+2(n)(n+1) + 2n+1 = 2n^2+4n+2 > 200$$
$$t = n^2 +2n+1 > 100$$
$t = (n+1)^2>100$
clearly $ t = 121$ is the smallest value of $t$.
There is an easier way to do this:
Note that the quadratic for $t$ can be written as $$t = (t-2r)^2$$. So $t$ must be a square number. The smallest square number greater than $100$ is $121$, and trying this, we see that $r$ is an integer. Thus $t = 121$ is the smallest possible value of $t$.
