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AnalogJoe · 05-26-2015, 09:06 PM

Thank you Dieter, im terribly sorry I didnt see this!

(05-26-2015 07:58 PM)Dieter Wrote:
(05-26-2015 07:36 PM)AnalogJoe Wrote: Im doing the following test program, my goal here is to find the roots of the following simple quadratic equation: x^2 + 5x + 1=0

OK. But that's not what you programmed:

(05-26-2015 07:36 PM)AnalogJoe Wrote: LBL 00
2
y^x
x<>y
5
x
1
+
RTN

There is a "+" missing after the "5 x". This way the function you provide is simply 5x + 1. And an additional x² that is left in Y.

Now guess where 5x + 1 becomes zero. ;-)

(05-26-2015 07:36 PM)AnalogJoe Wrote: The calculator works a little while and displays: -0.2

Got it now? ;-)
Simply provide the correct function, and you'll get what you want:

Code:

LBL 00 x² x<>y 5 * + 1 + RTN

BTW, do not use y^x as it is quite slow. There is a x² function that is shorter and much faster.

BTW2, "real cowboys" use Horner's method, for instance like this:

Code:

LBL 00 5 + * 1 + RTN

Looking at the equation you know that the roots will be negative, so try 0 and –5 as the two initial estimates.

Code:

0 ENTER -5 SLV 00 => -0,20871215252208... -2 ENTER -10 SLV 00 => -4,7912878474779...

There you are.

(05-26-2015 07:36 PM)AnalogJoe Wrote: Now the correct roots are -0.208712152 and -4.791287847

Another way of getting these is the dedicated quadratic solver SLVQ. It can be accessed via the X.FCN menu.

Code:

1 ENTER 5 ENTER 1 SLVQ => -4,791287847477920 x<>y => -0,2087121525220800

(05-26-2015 07:36 PM)AnalogJoe Wrote: So the solution is off the correct result. Also no matter what initial values I enter, I cant get the calculator to display something near -4.79 which is the second root.

You'll now know why: there is just one single root at exactly –0,2 for f(x)=5x+1. ;-)

Dieter