1. ## roots

Find the equation that roots inverted roots of the equation
$2x^{4}+9x^{3}+14x^{2}-9x+2=0$

2. Originally Posted by dapore
Find the equation that roots inverted roots of the equation
$2x^{4}+9x^{3}+14x^{2}-9x+2=0$
Say the roots of $2x^4+9x^3+14x^2-9x+2 = 0$ are $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$.
Is what you want the equation whose roots are $\frac{1}{\alpha_{1}}, \frac{1}{\alpha_{2}}, \frac{1}{\alpha_{3}}, \frac{1}{\alpha_{4}}$? See here...

3. Originally Posted by dapore
Find the equation that roots inverted roots of the equation
$2x^{4}+9x^{3}+14x^{2}-9x+2=0$
Just replace $\normalsize x$ by $1/x$ in the equation: $2x^{-4}+9x^{-3}+14x^{-2}-9x^{-1}+2=0$. Then multiply through by $x^4$. (The new equation is very similar to the original one.)

4. Originally Posted by Opalg
Just replace $\normalsize x$ by $1/x$ in the equation: $2x^{-4}+9x^{-3}+14x^{-2}-9x^{-1}+2=0$. Then multiply through by $x^4$.
Why does this work, may I ask?

5. Originally Posted by TheCoffeeMachine
Why does this work, may I ask?
If alpha is a solution, then $(x-\alpha)$ is a factor.

$\displaystyle\frac{1}{x}-\alpha=0\Rightarrow x=\frac{1}{\alpha}$

Do you see how the relation is the same as starting with $\frac{1}{\alpha}\mbox{?}$

6. I believe this works simply because $(\frac{1}{\alpha})^{-1}=\alpha$