# Question #256cc

Oct 26, 2016

$k = \frac{3 + \sqrt{9 + 8 N}}{2}$

#### Explanation:

$N = \frac{{k}^{2} - 3 k}{2}$

$\implies 2 N = {k}^{2} - 3 k$

$\implies {k}^{2} - 3 k - 2 N = 0$

We can now apply the quadratic formula with $a = 1 , b = - 3 , c = - 2 N$:

$k = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(- 2 N\right)}}{2 \left(1\right)}$

$\implies k = \frac{3 \pm \sqrt{9 + 8 N}}{2}$

Normally, we would be done. However, we are given that $k > 0$ and $N > 0$. As $N > 0$, we have

$N > 0$

$\implies \sqrt{9 + 8 N} > \sqrt{9} = 3$

$\implies 3 - \sqrt{9 + 8 N} < 0$

$\implies \frac{3 - \sqrt{9 + 8 N}}{2} < 0$

As we only want a positive value for $k$, we will dismiss the negative result, leaving us with a single answer:

$k = \frac{3 + \sqrt{9 + 8 N}}{2}$