# B Explanation for increase of degree of equation

1. Sep 21, 2016

### Mr Davis 97

So say that I have the quadratic equation $x^2 - 2x +1 = 0$. If I multiply each side by $x$, a valid operation, I get the new equation $x^3 - 2x^2 + x = 0$, which has a different solution set as the first, namely $\left\{ {0, 1}\right\}$. If I make the substitution $x^2 = 2x -1 = 0$, then I now have $x^3 - 2(2x - 1) + x = x^3 - 3x + 2$, which has the solution set $\left\{ {-2,1}\right\}$. I could repeat this process ad infinitum, but I am just wondering what is going on. If I am just doing valid operations and making valid substitutions, then why is the solution set constantly changing from what is originally was, i.e, $\left\{ {1}\right\}$?

Also, given that I transformed $x^2 - 2x + 1 = 0$ to $x^3 - 3x + 2 = 0$, is there any way to go in the reverse direction, that is, to start with $x^3 - 3x + 2 = 0$ and go to $x^2 - 2x + 1 = 0$?

2. Sep 21, 2016

### Staff: Mentor

The only change you're making is to enlarge the solution set.

If you leave your equations in factored form, it should be more obvious.
$(x - 1)^2 = 0$ Solution {1}
$x(x - 1)^2 = 0$ Solution {0, 1}
In your third equation, what you've done seems murky as you have written it, but if $x^2 = 2x - 1$, then $x^2 - 2x + 1 = 0$ or $(x - 1)^2 = 0$. Multiplying both sides by $(x - 1)^2$ raises the degree of the polynomial to 5, but there are no new values in the solution set.
Factor it.