Question

If the equation $a_n{x}^n+a_{n-1}{x}^{n-1}+\dots \dots +a_{1}x=0,a_1\ne 0,n\ge 2$, has a positive root $x=\alpha$, then the equation $n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\dots ..+a_1=0$ has a positive root, which is

• A. greater than $\alpha$
• B. smaller than $\alpha$
• C. greater than or equal to $\alpha$
• D. equal to $\alpha$

Answer 1

Answer: (B)

smaller than $\alpha$

=$\because a_n{x}^2+a_n+x^{n-1}+............+a_{1}x=0a_1\ne 0n\ge 2$

= has the root $x=\infty$

= $f^1\left(x\right)=x{a}_n{x}^{n-1}+(x-1)a_{n-1}{x}^{n-2}+.......a_n$

= $\because f\left(x\right)=0$

Let us take an example to see

Let a quadratic equation $x^2+2x-3=0$

$x^2+3x-x-3=0$

$x(x+3)-1(x+3)=0........\left(i\right)$

$x=1x=-3$

Now $f^1\left(x\right)=2x+1$

$f^1\left(x\right)=0=>x=-\frac{1}{2}..........\left(ii\right)$

From (i) and (ii) we can see that

The root of $f^1\left(x\right)$ is always less than the root of $f\left(x\right)$

Hence we can conclude

for $n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+......a_1$

has roots always less than $\alpha$ for the value of $\alpha$.