If, for a positive integer n , the quadratic equation, x(x + 1) +...

On simplifying we get the quadratic equations as
$\left(\mathop{\mathop{{x}^{2} + {x}^{2} + . . . + {x}^{2}}_{⏟}}_{n\text{times}}\right) + \left[1 + 3 + 5 + . . . + \left(2n - 1\right)\right]x + \left[1 . 2 + 2 . 3 + . . . + \left(n - 1\right)n\right] = 10n$
$n{x}^{2} + {n}^{2}x + \dfrac{n\left({n}^{2} - 1\right)}{3} = 10n$

${x}^{2} + nx + \dfrac{{n}^{2} - 31}{3} = 0$

Let, $\alpha, \beta$ are the roots of the above equation

$∴ \alpha + \beta = - n, \alpha\beta = \dfrac{{n}^{2} - 31}{3}$

Now, the difference of roots\left|\alpha - \beta\right| = 1$</p><p>$⇒ {\left(\alpha - \beta\right)}^{2} = 1$</p><p>$⇒ {\left(\alpha + \beta\right)}^{2} - 4\alpha\beta = 1$<br /><br />$⇒ {n}^{2} - \dfrac{4}{3}\left({n}^{2} - 31\right) = 1$<br /><br />$⇒ {n}^{2} = 121$<br /><br />$⇒ n = 11