Advanced Math/Summer math question
QUESTION: I need to find the roots of the polynomial given by ((x)^3)-4((x)^2)+((1)/((x)^2))-((x)^(4/5))-7. I have tried finding the roots via graphing calculator but I am not sure how to continue forward. Thanks!
ANSWER: Hi. Your expression is problematic mainly because of the exponent of 4/5, assuming I'm interpreting you notation correctly (your expression is also not technically a polynomial and I assume the expression should be set equal to 0 to make it an equation). Please carefully review your notation and I'll take a crack at it (otherwise, a whole lot of effort could be put into trying to come up with a solution to a typo).
Also, in addition to a graph solution, has your material discussed other techniques? Do you have any other examples from your work that are like the above expression?
---------- FOLLOW-UP ----------
QUESTION: In the original equation it was expressed as the fifth root of x to the fourth so I just changed it into exponential form for the sake of typing it out. We are supposed to use our graphing calculators and if our answers are irrational we are to round it to the thousandths place.
OK, a graphical "solution". This technique uses the fact that, even if you rearrange the terms in the equation, i.e., putting some on one side of the equal sign and some on the other, the value of x that solves it is still the same. If the 2 expressions on each side are plotted, then the point on the x-axis where they intersect is the value of x that solves the equation (which could be interpreted as the root of the original equation).
So take the equation (and I'm duplicating it here to make sure we're using the same expression)
x^3 - 4x^2 + 1/x^2 - x^(4/5) - 7 = 0
and rewrite as y = f(x) on each side
x^3 - 4x^2 = -1/x^2 + x^(4/5) + 7 = y
I plotted these 2 eqns in Excel (see attached image) and (after playing around a bit) came up with a value of x near 0.36 as their point of intersection. I assume you have a plotting application that can provide high resolution so that you can get the precision desired.
Anyway, you should check my arithmetic but this should show you how to do it. I should note that you need to play around a little to determine which pieces of the expression to plot against each other in order to focus in on the proper range of x values containing the solution. In fact, a little algebra to determine appropriate ranges of f(x) would be useful before just plotting away.