Discussion Forum

If $\alpha_{1},\alpha_{2},.......,\alpha_{n}$  are the roots of $x^{n}+px+q=0$,  then ($\alpha_{n}-\alpha_{1})(\alpha_{n}-\alpha_{2})...............(\alpha_{n}-\alpha_{n-1})=$

Answer:

Explanation:

We have,

  $x^{n}+px+q=0$,

 If   $\alpha_{1},\alpha_{2},.......,\alpha_{n}$    are roots of given equation so,

$x^{n}+px+q=(x-\alpha_{1})(x-\alpha_{2} )(x-\alpha_{3}).......(x-\alpha_{n})$

$\Rightarrow$    $\frac{x^{n}+px+q}{x-\alpha_{n}}=$  $(x-\alpha_{1})(x-\alpha_{2} )(x-\alpha_{3}).......(x-\alpha_{n-1})$

$\therefore$  $\lim_{x \rightarrow \alpha_{n}}\frac{x^{n}+px+q}{x-\alpha_{n}}=\lim_{x \rightarrow \alpha_{n}}$ $(x-\alpha_{1})(x-\alpha_{2}) (x-\alpha_{3}).......(x-\alpha_{n-1})$

$\lim_{x \rightarrow \alpha_{n}}\frac{nx^{n-1}+P}{1}=(\alpha_{n}-\alpha_{1})(\alpha_{n}-\alpha_{2})......(\alpha_{n}-\alpha_{n-1})$

 $\Rightarrow$      $n \alpha_{n}^{n-1}+p=(\alpha_{n}-\alpha_{1})(\alpha_{n}-\alpha_{2})......(\alpha_{n}-\alpha_{n-1})$