Question

Show that the equation
$x^n+n{x}^{n-1}+n(n-1)x^{n-2}+....+|\underset{–}{n}=0$ cannot have equal roots.

Answer 1

$x^n+n{x}^{n-1}+n(n-1)x^{n-2}+....+|\underset{–}{n}=0$

Consider $f\left(x\right)=x^n+n{x}^{n-1}+n(n-1)x^{n-2}+....+|\underset{–}{n}$

$f\left(0\right)=|\underset{–}{n}\ne 0$ - Therefore $x=0$ is not a root of $f\left(x\right)=0$

$f^{\prime }\left(x\right)=n{x}^{n-1}+n(n-1)x^{n-2}+....+|\underset{–}{n}$

Assuming that $f\left(x\right)=0$ has $\alpha$ as equal roots, then $(x-\alpha )$ will be a factor of both $f\left(x\right)$ and $f^{\prime }\left(x\right)$. Therefore,

$f\left(\alpha \right)={\alpha }^n+n{\alpha }^{n-1}+n(n-1){\alpha }^{n-2}+....+|\underset{–}{n}=0$

Also, $f^{\prime }\left(\alpha \right)=n{\alpha }^{n-1}+n(n-1){\alpha }^{n-2}+....+|\underset{–}{n}=0$

$f\left(\alpha \right)–f^{\prime }\left(\alpha \right)={\alpha }^n=0$

$⟹\alpha =0$

But $x=0$ it not a root of $f\left(x\right)=0$, therefore $f\left(x\right)=0$ cannot have equal roots.

Hence Proved