If the equation [a_nx^n + a_n - 1x^n - 1 + ...........

Question

If the equation $${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$

$${a_1} \ne 0,n \ge 2,$$ has a positive root $$x = \alpha $$, then the equation

$$n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ........... + {a_1} = 0$$ has a positive root, which is

Accepted Answer

Let $$f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$

The other given equation,

$$na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)$$

Given $${a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0$$

Again $f(x)$ has root $$\alpha ,\,\,\, \Rightarrow f\left( \alpha  \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,f\left( 0 \right) = f\left( \alpha  \right)$$

$$\therefore$$ By Roll's theorem $f'(x)=0$  has root be-

tween $$\left( {0,\alpha } \right)$$

Hence $f'(x)$ has a positive root smaller than $$\alpha .$$

Solutions · 1