# Concept problems regarding equations of higher degree

1. Nov 14, 2009

### jeremy22511

1. The problem statement, all variables and given/known data
We generally accepted in lower forms, that when we deal with equations of higher degree, arriving at solutions which need to be rejected is normal, even natural.
However, now that I think of it, doesn't arriving at wrong answers mean only that our deductive argument contains flaws? Otherwise, why else would we have to reject some answers?
Thanks.

J

2. Relevant equations

3. The attempt at a solution

2. Nov 14, 2009

### HallsofIvy

That depends upon what you mean by "flawed". If a "deductive method" gives the right answer, it isn't flawed, is it? I think what you are talking about is when we, say, square both sides of an equation to get rid of a square root, or multiply both sides of an equation by something to get rid of fractions. When you do that, you do change the equation to something that is not exactly equivalent. But as long as you are aware that you are doing that, and take steps to make sure it doesn't give you a wrong answer, as, for example, checking any answers, there is no flaw in your "deductive argument".

As long as we multiply by constants, we know that a number is a solution to our original equation if and only if it is a solution to our new equation. If we multiply both sides of the equation by something involving the unknown variable, then any solution to our original equation must be a solution to the new equation but not vice-versa.

If our original equation is f(x)= g(x) and we multiply both sides by h(x) to get f(x)h(x)= g(x)h(x), then any solution to our original equation is still a solution to our new equation but not necessarily the other way. Any solution to h(x)= 0 is also a solution to f(x)h(x)= g(x)h(x).

When we multiply both sides of an equation by a constant, c, our "deductive argument" is "x is a solution to f(x)= g(x) if and only if it is a solution to cf(x)= cg(x)."

If our "deductive argument" were "x is a solution to f(x)= g(x) if and only if it is a solution to f(x)h(x)= g(x)h(x)", then it would indeed be flawed.

But it is not. Our argument is "any solution, x, to f(x)= g(x) is a solution to f(x)h(x)= g(x)h(x)". That is perfectly correct. It is the other way, "any solution to f(x)h(x)= g(x)h(x) is a solution to f(x)= g(x)" that is incorrect. And we don't claim that. That's why we have to specifically check solutions.

3. Nov 14, 2009