In a previous post talk about solving equations with objective function, this allowed us to calculate a value that satisfied the equation f (x) = 0, but an equation can have not only positive but also a solution some negative. In this post I will explain how to obtain a root positive and one negative in an equation using the Excel Solver.
Consider the following graph:
This graph is plotted the polynomial f (x) = x ^ 4 - 2 * x ^ 3 - 16 * x ^ 2 - 2 * x - 15
The idea is to find a couple of points x1 \u0026lt;0 and x2> 0 such that f (x1) = 0, f (x2). Graphically we see that in real numbers such points if there are, As if using the solver find a solution. In the example I assume you already have the solver suitable for use, but so you can use this document to install Solver.
I created the following function for f (x) so that it easier to evaluate the polynomial.
Find the first value, x1 negative. Suppose in Cell E20, we have x1 = -2.5 value in E21 evaluate the function for this value the result is -39.7. Then the cell that should vary is x1, talque f (x1) = 0 and meets x1 \u0026lt;0. Using the solver, it would be like this (listed Solver Data):
1. Indicate the cell to vary and the objective function:
2. Click the Add button to stable condition x1 \u0026lt;0, select E20
3. Accept and resolve.
Likewise, you can get the solution for x2> 0. You can download the help file here.
Greetings.
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