On my Casio scientific calculator I am able to quickly solve equations in the form of ax^3+bx^2+cx+d=0 (or with a degree of two) for x by simply entering the a,b,c, and d values in a table. I've done a decent amount of searching and the closest thing I've gotten is putting the equation into 'solve symbolic view', then pressing Num and setting y=0. That seems to only give me the positive x-intercept, though. Any suggestions?
(02-17-2016 12:32 PM)MattH Wrote: [ -> ]On my Casio scientific calculator I am able to quickly solve equations in the form of ax^3+bx^2+cx+d=0 (or with a degree of two) for x by simply entering the a,b,c, and d values in a table. I've done a decent amount of searching and the closest thing I've gotten is putting the equation into 'solve symbolic view', then pressing Num and setting y=0. That seems to only give me the positive x-intercept, though. Any suggestions?
Have you tried PROOT?
Is that anything like what David Hayes uses in this video to find the x-intercepts? (He uses the
root() function after putting the equation into the
Function app.)
https://www.youtube.com/watch?v=pDIDmq1Y5Pc If not, where do I find Proot? I've tried the root function and it gives me a root, but I have to supply a guess which I could see being a pain at times.
Also, on page #32 of
http://h20331.www2.hp.com/hpsub/download...ionsv3.pdf the author shows how to use several different CAS options like
zeros or
solve. These methods return an empty pair of brackets for me but they seems to work for him. He is in spreadsheet app in the backround, does that matter? I'm going to keep messing with it.
Ideally, I'd like something that doesn't require me to input the variables, powers, and + signs, and instead just gives me a template to work with.
Also: Toolbox->CAS->Polynomial->Find Roots
You give it a vector with your values, out pops the roots (with complex roots also).
Didier, what's the easiest way to enter those values? Do I create a matrix? I've tried using the [[]] template and deleting a set of brackets but it just returns an empty set of brackets. Also is the equation in the the form x^4 +2x^3 -25x^2 -26x +120=0?Finally, do you know of a way to quickly select the proot() command without searching through that entire list of commands in the toolbox catalog?
Tim, can you give me an example of how I would do what you suggested? If I wanted to find the roots for the above equation, how would I input it inside the parenthesis?
Sorry for the barrage of questions but my math knowledge is probably lacking compared to you guys and this is my first graphing calculator since a TI-83 in high school over 10 years ago.
I came across Carl's tutorial on CAS solve() which has really helped me out.
https://docs.google.com/document/d/1eTfT...OffKb4/pub
Only problem is that I'd really like to use the proot() command but when I do I get a really weird answer compared the the x-intercepts that Carl gets when he does it in example 2 of the tutorial. Any idea why that is?
I put a screen capture in an attachment.
(02-18-2016 02:19 AM)MattH Wrote: [ -> ]using the [[]] template
That will be a matrix template. To do a simple vector, just press SHIFT-5. Press the [,] key to move to the next item. Note that if you accidentally made a matrix and the cursor is inside the matrix and there is no more then 1 row), just press SHIFT-5 and it will toggle it back to a vector.
proot([1,2,-25,-26,120]) --> [-5.,-3.,2.,4.]
Quote:Finally, do you know of a way to quickly select the proot() command without searching through that entire list of commands in the toolbox catalog?
Toolbox->CAS (button on the bottom of the screen)->Polynomial->Find Roots
Awesome! Thanks Tim! I've actually got one more issue that you can find just above your last post. I'm curious if you have any idea what's going on with it.
Edit: Well I figured out the problem was that I wasn't using the vector method, which is kind of weird because in Carl's pictures he wasn't using it either. I'm still kind of curious what exactly the calculator is solving when I just put the coefficients directly inside the parenthesis.
I'm stoked to have finally learned how to do this quickly though, thanks for all the help guys. I'm loving this calculator more and more every day. It's like this little pretty little rectangle full of secrets.
While I've got you here Tim, I had a few related questions if you don't mind.
First, what exactly does the "x t theta n" button do? Most of the time it seems to give me an 'x', but a few minutes ago it was giving me an 'a' while in CAS, but now it stopped. Is there a way to get it to input different characters?
Second, what is the purpose of the 'plus or minus' symbol inside the vector brackets?
(02-18-2016 07:35 AM)MattH Wrote: [ -> ]First, what exactly does the "x t theta n" button do? Most of the time it seems to give me an 'x', but a few minutes ago it was giving me an 'a' while in CAS, but now it stopped. Is there a way to get it to input different characters?
Second, what is the purpose of the 'plus or minus' symbol inside the vector brackets?
For a quick definition of key function, push [Help] and the [Keys] soft key. It will ask you to "Press the key you want help on." Very handy on-calc reference!
-Dale-
(02-18-2016 07:35 AM)MattH Wrote: [ -> ]Second, what is the purpose of the 'plus or minus' symbol inside the vector brackets?
The +/- symbol in vectors/matrices permits you to add a column/row. It's a sort of placeholder when you navigate in the vector/matrix with the cursor or finger.
When you separate the values with comma it does nothing.
(02-18-2016 03:28 AM)MattH Wrote: [ -> ]I came across Carl's tutorial on CAS solve() which has really helped me out.
https://docs.google.com/document/d/1eTfT...OffKb4/pub
Only problem is that I'd really like to use the proot() command but when I do I get a really weird answer compared the the x-intercepts that Carl gets when he does it in example 2 of the tutorial. Any idea why that is?
I put a screen capture in an attachment.
That was an earlier version of the HP Prime software - I guess didn't require [1,-2,-5,24] brackets at that time.
Great, thanks for answering all my questions.