need help with solving some simple equations on calculator...

[MrPolite]

I have some equations that are similar to this:
(5x!) + 3x^2 - 25347 = 0 (answer: x=7)

I need to know how to solve something like this using a TI-83 calculator. I would really appriciate if someone could tell me how to do this. (umm, ahem, does it need a calculator at all? )

[sql_lall]

probably the easiest way to do this is graph
y=(5x!) + 3x^2 - 25347
then find intercepts.

However, there is the 'solver' thing, but i'm not sure if it copes with factorials.

[Allegro]

Hello Mr Polite,

None of the suggestions of sql_lall worked on my TI-83,so I tried a different approach.
I made a program called "FAC" , and the listing is as follows:

Prompt X
5X!+3X²->A
Disp"A=",A

You start with an integer , for instance 3, then 4 and so on.
You will find f(7)=25347.

[MrPolite]

Originally posted by Allegro
Hello Mr Polite,

None of the suggestions of sql_lall worked on my TI-83,so I tried a different approach.
I made a program called "FAC" , and the listing is as follows:

Prompt X
5X!+3X²->A
Disp"A=",A

You start with an integer , for instance 3, then 4 and so on.
You will find f(7)=25347.

oh so you're basically testing it? hmm good idea, I guess that works

btw an easier way to test would be to put the equation in Y1 (or Yn). then in the main section go to Vars and select Y1, so you can test numbers by entering Y1(7)

thanks, I guess I would just use this testing method, sounds like it works faster than any other method

[sql_lall]

Yeah, it might not have looked good cos you can't have factorials of non-integers or decimals with more that x.5

However, to graph it nicely, instead of "X" in my previous equation, put ipart(X). Set Xmin = 0, XMax = 10, Ymin=-10, YMax=10 and you get a wierd step-like graph, intersecting when ipart(X) = 7
=> X=7 is the solution.

[da_silvy]

Use solver...

Math Button >> Solver

Enter the equation, then alpha enter (--> solve)

Worked for me on a TI-83+

[RAEsquivelC]

Of course, 5! = 120, and the equations gives, for x:

120-25347+3*x^2=0
3*x^2 = 25227
x^2 = 8409
x=91.7006, NOT 7.

The general form of this equation would be:
ax^3 + b = 0, and its solution would be:
x=(-b/z)^(1/2)

So, on my simple calculator, once I know the above, putting
b=-24347+5!
a=3
I would enter:
5! and get 120
-25347= and get -25227
change sign and get 25227
/3= and get 8409
sqrt and get 91.70059978 (My calculator gives ten significant figures).

[sql_lall]

Nice points
however, not sure if you misread the Question
it is (5x)!, not 5!

Also, just a question about your lines:
"ax^3 + b = 0, and its solution would be:
x=(-b/z)^(1/2)"

1) what is z?
2) wouldn't it be just (-b/a)^(1/3)

[RAEsquivelC]

sql_lall, thank you for noticing the error of my ways!

My first error was in mistaking (5x)! for (5)!, which makes the rest of my analysis totally unsuitable.

My second error was a typo, I pressed the "z" key, instead of the "a" key.

My third error was in showing the power as 1/2, when, as you point out, it is 1/3.

And, my fourth-and most grievious error-was not checking!!!

Since I think various of you have already given good solutions to the problem, I'll just bow out of this one.

[sql_lall]

No worries.
I mean, you put the effort in to work stuff out, more than a lot people around here