Solving an interesting polynomial with degree 4?

Question

So the equation is as follows:

$$ 6x^2 -\ 25x \ + 12 \ +\ \frac6{x^2}\ + \frac{25}{x} = 0$$

So one thing that is immediately observable is that pairs of roots will be of the from $$x_1=-\frac{1}{x_2}$$

How do I go about solving this equation systematically ? I have already solved it based on trial error to get roots as 2,3,-1/2,-1/3. But I need to show systematic manipulations to get there.

Thanks a lot !

Could you please double-check that the equation you typed here matches the equation you're working with? Did you mean $25x$ instead of $25x^3$? And is there the right number of minus signs? — Chris Culter, Feb 07 '19 at 06:27
@ChrisCulter - Yeah edited it. The number of minus signs is correct. — DS112, Feb 07 '19 at 06:30
@LordSharktheUnknown - As in ? Substitute y instead of x in the above equation ? wouldn't that unnecessarily complicate things ? — DS112, Feb 07 '19 at 06:36
@DS112: No, use the fact that $y^2=x^2+1/x^2-2$ to express your original polynomial as a combination of $y^2$, $y$, and $1$. — hmakholm left over Monica, Feb 07 '19 at 06:37
Possible duplicate of Solve for $x$ in $6x^2-25x+12+\frac{25}{x}+\frac{6}{x^2}=0$ — Martin R, Feb 07 '19 at 06:54
@HenningMakholm - Got ya. Solved and got the right answers. But how did you reach the epiphany that this is the right substitution to make ? — DS112, Feb 07 '19 at 06:58

score 2 · Answer 1 · answered Feb 07 '19 at 06:43

2

Hint: Write $$6\left(x^2+\frac{1}{x^2}\right)-25\left(x-\frac{1}{x}\right)+12=0$$ and substitute $$t=x-\frac{1}{x}$$

answered Feb 07 '19 at 06:43

Dr. Sonnhard Graubner

95,283

3

I think $$6\left(x^2-2+\frac{1}{x^2}\right)-25\left(x-\frac{1}{x}\right)+14=0$$ is better. – Arthur Feb 07 '19 at 06:52
2

@Arthur: And $+24$ would be better yet (notice the factor of $6$ in front!) – hmakholm left over Monica Feb 07 '19 at 07:33

Solving an interesting polynomial with degree 4?

1 Answers1