The roots of the equation $\left|\begin{array}{ccc}{ }^{\mathrm{x}}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1}\\ { }^{\mathrm{x}+1}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}\\ { }^{\mathrm{x}+2}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}-1}\end{array}\right|=0$ are

$\left|\begin{array}{ccc}{ }^{\mathrm{x}}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}\\ { }^{\mathrm{x}+1}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}}\\ \mathrm{x}+2{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}}& { }^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}}\end{array}\right|=0$           (1)

$\Rightarrow \left|\begin{array}{ccc}\frac{\mathrm{x}!}{\mathrm{r}!(\mathrm{x}-\mathrm{r})!}& \frac{(\mathrm{n}-1)!}{\mathrm{r}!(\mathrm{n}-\mathrm{r}-1)!}& \frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}\\ \frac{(\mathrm{x}+1)!}{\mathrm{r}!(\mathrm{x}+1-\mathrm{r})!}& \frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}& \frac{(\mathrm{n}+1)!}{\mathrm{r}!(\mathrm{n}-\mathrm{r}+1)!}\\ \frac{(\mathrm{x}+2)!}{\mathrm{r}!(\mathrm{x}+2-\mathrm{r})!}& \frac{(\mathrm{n}+1)!}{\mathrm{r}!(\mathrm{n}+1-\mathrm{r})!}& \frac{(\mathrm{n}+2)!}{\mathrm{r}!(\mathrm{n}-\mathrm{r}+2)!}\end{array}\right|=0$

Taking $\frac{\mathrm{x}!}{\mathrm{r}!(\mathrm{x}-\mathrm{r})!}$ common from C₁, we have quadratic equation in x.

Now in (1), if we put x = n - 1, C₁ and C₂ are the same; hence, x = n - 1 is one root of the equation.

If we put x = n, then C₁ and C₃ are same. Hence, x = n is the other root.