+128731 x^4-x^3=500
The answer that Anonymous gave, (5), IS correct. The question is....how did we arrive at that ??
Rearranging by subtracting 500 from both sides, we have
x^4-x^3 - 500 = 0
We have a theorem in math known as the "rational zeros" theorem that says that IF we have any "rational" solutions to the above (solutions that can be expressed as a fraction), these solutions come from all the possible factors of 500 divided by all the possible factors of the coefficient in front of the x^4 term, i.e., "1."
Of course, listing all the possible factors of 500 would be lengthy, but we can list the first few...we have ±1, ±2, ±4, ±5, etc. and all the possible factors of 1 are just ±1.
So..our list of (possble factors of 500)/(possible factors of 1) would include ±1, ±2, ±4, ±5, etc.
And we would find, by substituting these for "x' - one at a time, of course - into the above equation would provide us with our (only) "rational" solution of x = +5. Note that +5 is "rational," because we can express it as (+5)/(+1).
Note one thing......the "rational zeros" theorem doesn't guarantee that we have any "rational" solutions....it just says that, if we do, they will come from the list of (all possible factors of 500)/(all possible factors of 1).
(Another theorem called the "bounds on zeros" theorem will tell us when we can stop searching for any additional zeros.)
+33616 There is in fact another solution on the real number line (approximately -4.4972), but it is irrational (i.e. can't be expressed as a ratio of two integers). This can be seen easily by plotting a graph of x^4-x^3 against x. There are also two complex number solutions.
