Solved a quartic equation by dividing

Question

I was asked to solve: $$x^4+2x^3-22x^2+2x+1 = 0$$ Without using differential calculus (Newton's Method). My Progress: Dividing by $x^2$, I get: $$x^2+2x-22+\frac{2}{x}+\frac{1}{x^2} = 0$$ $$x^2 +\frac{1}{x^2} +2x +\frac{2}{x} -22 = 0$$ $$(x+\frac{1}{x})^2 +2(x+\frac{1}{x}) - 24 = 0$$ let $k = (x+\frac{1}{x})$ $$k^2+2k-24 = 0$$ $$(k+6)(k-4) = 0$$ $k = -6$ or $k = 4$

Putting $4 = x + \frac{1}{x} $ leads to $x^2 − 4x + 1 = 0$ as above, and putting $−6 = x + \frac{1}{x}$ leads to $x^2 + 6x + 1 = 0.$

Never mind the answer that I get after that. The big question that I have is that I divided by $x^2$ at the top. My maths teachers always told me that one should never divide by a variable but I did it and got 4 different roots for this question. Why am I not allowed to divide by a variable and when is it possible that I can divide by a variable? It made me solve this equation.

http://math.stackexchange.com/questions/480102/quadratic-substitution-question — lab bhattacharjee, Sep 24 '14 at 07:35
I think it's legitimate to divide either sides of an equation anything non-zero. — lab bhattacharjee, Sep 24 '14 at 07:38
You should first check if zero is a solution or not. Then you can divide by all non-zero root $x$. If you found that zero was a solution, you have to add it to the set of roots. — Tom-Tom, Sep 24 '14 at 07:49
Since $x=0$ is not a solution of the equation, I think that the division by $x^2$ is perfectly legitimate. — Claude Leibovici, Sep 24 '14 at 07:55
I think that you have a sign error at the second line. $$x^2+2x-22+\frac{2}{x}+\frac{1}{x^2} = 0$$ should conform to the first line. — Claude Leibovici, Sep 24 '14 at 07:59
If the first equation is correct, you still have sign errors in many places. — Claude Leibovici, Sep 24 '14 at 08:19

Peter Huxford · Accepted Answer · 2019-03-13T01:28:03.047

The reason why, in general, it is considered bad practice to divide by a variable, is because you may lose out on solutions, when whatever you divide by happens to be zero. It is however perfectly valid if you can ensure you are dividing by a nonzero quantity.

For example: solving $x(x-1)=x(2x-3)$. Naively one might divide by $x$ on both sides, and get that $x-1=2x-3\implies x=2$. However they would be implicitly assuming that $x\neq0$ (as you cannot divide by zero). One way to deal with this, for this example, is to bring all terms onto the same side and factor (which is usually prefered), however it would be equally valid to consider two cases instead.

Case 1: $x=0$. In this case, the equation is satisfied, so we have a valid solution.

Case 2: $x\neq0$. In this case, we can divide both sides by $x$, because it is nonzero, and proceed as before.

So in general, if it is inconvenient to factor, you can just consider the two cases when what you want to divide by is zero, and when it is nonzero. In your case, it is pretty easy to check that $x=0$ does not satisfy the equation, so it is perfectly valid to divide by the variable $x$ in this case, without any additional casework.

How about solving it? That equality never holds for real numbers $a,x$, if you subtract the right side from the left you get $\frac{x-a}{x-a}=1$, so the equation becomes $1=0$, which is false. — Peter Huxford, Sep 24 '14 at 07:47
Assume that there was a solution. Then by my working 1 would be equal to 0. This is clearly false. So there cannot be any solutions. — Peter Huxford, Sep 24 '14 at 08:04

score 0 · Answer 2 · answered Oct 17 '14 at 00:01

The teacher is half right and half wrong. First you must check if the constant term in the polynomial equals zero. If so, then zero is a root and you can factor a general degree of x^n where n is a natural number. Then you can solve both factors. If the constant term is not zero, then zero cannot be a root, like in your example where the constant term equals one. In this case you can divide the equation by the variable without any problems with undefined expressions. The teacher just wanted to make things simpler for the rest of the class, yet your solution is mathematically valid. Please tell the teacher about what I typed.

Solved a quartic equation by dividing

2 Answers2