An alternative method to solve inequalities of Nth degree

I remember when I was in Middle School, it was a pain for other student to solve the quadratic and cubic inequalities. What the textbook taught is what I seen as a stupid and time-wasting method: Just like solving the quadratic and cubic inequalities, we first need to need all the terms in one side so we factorize it into their factors; then they list the factor and individually determine the signs of each factor. The stupidity lays in the need to determine the sign of each factor individually, I devise a faster method at that time.

I was thinking at that time, since all these factors are related, why must be they be deal with individually? Obviously, since all of them are referring to a single variable x, i.e. the signs of each factor are not independent, why we can’t take advantage of their inter-relationships? What is the implicit inter-relationship between the factors? If we transform all of the factor into comparable form, then we can list them in ascending or descending sequences like this: x+1>x-3>x-3/4… etc. Now, clever reader may already see the trick I play here: The central idea is we can save ourselves a few steps because we can’t have the case which a Greater factor being negative while a Smaller factor being positive. Thus we can list all feasible (and logical) case for the signs of the factors to be determined instead of blindly list ALL THE POSSIBLE case.

To save us more step, we know that we require an odd number of negative factor to resulted into negative, and we require an even number of negative factor to resulted into positive. And the result of multiplication of all positive factor must resulted into positive number.

Thus, we can draw the boundary somewhere in the factors. Assume one factor is being positive, then any factor which is greater than that factor would always be positive; and any factor smaller than that factor would always be negative. Thus, in the case of x+1>x-3>x-3/4, if we assume X+1>0, then we must have x-3<(x-3/4)<0, therefore x>-1 is the solution for (x+1)(x-3)(x-3/4)>0; now if we assume x-3>0, then we must have x+1>0 and x-3/4<0, since we have an odd number of negative factor so the result of multiplication must be negative; therefore x>3 as the solution for (x+1)(x-3)(x-3/4)<0. I think we can skip the step which we assume x-3/4>0, since then we must have x+1>x-3>x-3/4>0, therefore x>3/4 is the solution for (x+1)(x-3)(x-3/4)>0. Similarly, if we assume x+1<0, then we must have 0>(x-3)>(x-3/4), therefore x<-1 is the solution for (x+1)(x-3)(x-3/4)<0.

Using this methodology, an inequality of n-th degree only need to be evaluate n+1 times at maximum instead of evaluating 2^n times. Plus, using logical deduction with the axioms which only odd number of negative factor will result in negative number; and only even number of negative factor would guarantee a positive number, we can further reduce the cases which we need to evaluate. For instance, if we want to solve a  seven factors inequality which is greater than zero, we can only cases of 2 negative, 4 negative, 6 negative factors and no negative factors.

~ 由 newnewhkcc1976 於 1 六月, 2008.


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