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March 11th, 2008, 06:30 PM  #1 
Newbie Joined: Mar 2008 Posts: 5 Thanks: 0  Programming TI83 calculator to Solve Quadratic Equation?
Does anyone know how to program the TI83 calculator to solve for quadratic equations? Thank you in advance for any input.

March 11th, 2008, 09:54 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
If you just want numerical solutions that's easy. Just have it display the results with + and with . If you want it in simplified symbolic form... well, I did this on my TI83 ten years ago, but it's a lot of coding for such a simple language.

March 12th, 2008, 08:38 AM  #3 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]Hi: For ax^2 + bx + c = 0, program as, :input a :input b :input c :display 0.5[b + sqrt(b^24ac)] :display 0.5[b  sqrt(b^24ac)] : Regards, Rich B. rmath4u2@aol.com[/color] 
March 13th, 2008, 01:20 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,712 Thanks: 1806 
You might get caught out (if you do it that way) by a sufficiently illconditioned equation.

March 13th, 2008, 10:35 AM  #5 
Member Joined: Feb 2008 Posts: 89 Thanks: 0  
March 13th, 2008, 11:43 AM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
nikkor, you forgot to divide by a. Quote:
a = (16) b = (33) c = (16) b^2 = (10)00, rounded from 1024 ac = (25)0, rounded from 256 4ac = (10)00 b^2  4ac = (0.0) sqrt(b^2  4ac) = (0.0) b + sqrt(b^2  4ac) = (33) 2a = (32) (b + sqrt(b^2  4ac))/(2a) = (1.0) which compares reasonably to 0.(78), the best result expressible in the system, considering that 510% precision can be lost per operation in a system with this few digits. Practically, with as many digits as a TI83 has, what's the worst case? I'm assuming this will be for algebra or calculus, not numerical analysis.  
March 13th, 2008, 03:10 PM  #7 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]You are indeed correct; I forgot the 'a'. Regarding rounding error, I suppose in the interest of accuracy, the program should report two separate terms for each root as, :b/2a :display "+ square root":(b^24ac)/(4a^2) :b/2a :display " square root":(b^24ac)/(4a^2) : Assuming rational coefficients, accuracy should be covered and rounding left to the discretion of the program user. What do you think? Regards, Rich B.[/color] rmath4u2@aol.com 
March 14th, 2008, 05:00 AM  #8  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
 
March 15th, 2008, 05:28 AM  #9  
Global Moderator Joined: Dec 2006 Posts: 19,712 Thanks: 1806  Quote:
 
March 15th, 2008, 09:11 AM  #10  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
If a class requires so much precision that that much subtractive cancellation is a problem with the TI83's normal working precision, then the techniques for handling it should be covered in the class itself. I get the impression that this is more of a high school class than a graduate class. My solution to that (non)problem in high school was, as I wrote above, to evaluate the values symbolically. But this expands the program from 46 lines to 50100 lines, or several hundred with good corner case handling and aligned displays.  

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