• Mar 9th 2011, 08:52 AM
lucas7
Hello, I'm having problems to solve this equation:

$6x^2-5x^-^1+1=0$

I really don't know how to do.

I don't know if it will be $5.\frac{1}{x}$ or $\frac{1}{5x}$ and how do I get a b and c to make the use bhaskara formula.

• Mar 9th 2011, 02:01 PM
earboth
Quote:

Originally Posted by lucas7
Hello, I'm having problems to solve this equation:

$6x^2-5x^-^1+1=0$

I really don't know how to do.

I don't know if it will be $5.\frac{1}{x}$ or $\frac{1}{5x}$ and how do I get a b and c to make the use bhaskara formula.

1. As far as I'm informed Bhaskara's formula is used to solve quadratic equations. After some transformations your equation will be a cube equation:

$6x^2-5 \cdot \dfrac1x + 1=0~\implies~6x^3+x-5=0$

If you don't want to use numerical methods you have to use Cardano's formula.

2. If and only if the original equation is

$6x-5 \cdot \dfrac1x +1=0~\implies~6x^2+x-5=0$

Then you can use Bhaskara's formula.
• Mar 10th 2011, 01:41 AM
lucas7
Well I got to the $\frac{6x^3-5+x=0}{x}$ point. But I do not understand how to reduce it to x^2 so I can use the bhaskara forum and solve it myself. Could you or someone please explain it? Thanks so much in advances