You may use the symbolic "solve" command to solve these equations in Mathcad.
Or you may use a solve block with "find"
Or you may use the "root" function.
Look them up in the help, give one of them a try and if you experience problems then come back here and post your worksheets with your attempts.
I would advise against using the symbolic "solve". In case of your first function it would return only the trivial solution x=0.
Using one of the numeric methods you get the other two solutions as well if you provide appropriate guess values.
If you really need it to be done using the simple bisection method for trainings purposes in your education, you have to look up the method in your lecture notes. You should not need Mathcad to do so other than for plotting to find suitable start values and to evaluate the functions numerically to see where the sign changes.
Of course you could write a Mathcad function wich takes the function and two initial values as its arguments and returns a list of left and right boundary values (until a certain precision is achieved). If this is your goal, then make a start, give it a try and if you get stuck, come back here and post your worksheet with your attempts.
@ppal wrote:
Depends on your tolerance settings -
No, it depends on the math and the two zeros are close, but they are definitely different. In other words - the graphs of the two functions have a couple of common points but none of them lies exactly on the x-axis.
The default tolerance setting (10^-3) is all thats necessary to get the two different solutions mentioned.
0.5 is best we can get to.
???
Try to use a solve block with both function forced to be zero for the same x value. It will fail for good reason (even Mathcads error message is wrong and not helpful as is unfortunately usual with solve blocks.
Here in detail:
I guess the question here is about two different tasks of homework ...
