Difference between revisions of "TI-BASIC:Quadratic Formula"

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The ordinary quadratic formula, and its alternate form are used:
 
The ordinary quadratic formula, and its alternate form are used:
  
[[TI-BASIC:Math|Math]]
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<math>
 
\begin{align}
 
\begin{align}
 
\mathrm{For} \hspace{5pt} ax^2+bx+c = 0, \\ \\
 
\mathrm{For} \hspace{5pt} ax^2+bx+c = 0, \\ \\
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x = \frac{2c}{-b\pm\sqrt{b^2-4ac}}
 
x = \frac{2c}{-b\pm\sqrt{b^2-4ac}}
 
\end{align}
 
\end{align}
[[TI-BASIC:/math|/math]][[Category:TI-BASIC]]
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</math>
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[[Category:TI-BASIC]]
 
[[Category:TIBD]]
 
[[Category:TIBD]]

Revision as of 19:25, 24 February 2016

File:@@

Routine Summary

Solves for the complex roots of a quadratic equation.

Inputs

A, B, C - the constants in Ax^^2^^+Bx+C=0

Outputs

Ans - a 2-element list of the two roots (or the one repeated root)

Variables Used

A, B, C

Calculator Compatibility

TI-83/84/+/SE Author

thornahawkAuthors

@@URL: United TI Download

file quadraticformula.zip

-2C/(B+{-1,1}√(B²+4ACi²

For optimization purposes, an alternate form of the quadratic formula is used (see below for the formulas). The {-1,1} list is used to replace the ± symbol -- lists can be used just as well as numbers in expressions, making the result a list as well. By using i² in the expression in place of subtraction, the routine avoids having to activate A+bi mode to allow complex roots.

The output is a complex list in Ans of the two roots. If there is only one root, it will be returned twice. If both roots are real, they will still be returned correctly, but stored internally as complex numbers -- so use the real( command on them if you want to pass them to commands that don't accept complex arguments.

Advanced

The ordinary formula above can give poor results if B is much larger than A and/or C. In that case, an alternate routine can be used:

√(B²+4ACi²
If 0>real(Ansconj(B
-Ans
-.5(B+Ans
{Ans/A,C/Ans

Formulas

The ordinary quadratic formula, and its alternate form are used:

<math> \begin{align} \mathrm{For} \hspace{5pt} ax^2+bx+c = 0, \\ \\ x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x = \frac{2c}{-b\pm\sqrt{b^2-4ac}} \end{align} </math>