On Thu, 6 Nov 2008 22:51:35 -0800 (PST), stndby
wrote:
>3.x + 5.x^2 + 34.x^3 + 24.x^9 = 34234
>
>how can i solve like this questions...can anybody help ??
On HP 50g: [RightShift] [NUM.SLV] and solve for x.
Damir
This is a discussion on x + x square equal any number ??? - Hewlett Packard ; 3.x + 5.x^2 + 34.x^3 + 24.x^9 = 34234 how can i solve like this questions...can anybody help ??...
3.x + 5.x^2 + 34.x^3 + 24.x^9 = 34234
how can i solve like this questions...can anybody help ??
On Thu, 6 Nov 2008 22:51:35 -0800 (PST), stndby
wrote:
>3.x + 5.x^2 + 34.x^3 + 24.x^9 = 34234
>
>how can i solve like this questions...can anybody help ??
On HP 50g: [RightShift] [NUM.SLV] and solve for x.
Damir
> 3.x + 5.x^2 + 34.x^3 + 24.x^9 = 34234
>
> how can i solve like this questions...can anybody help ??
[ 24 0 0 0 0 0 34 5 3 -34234 ] PROOT
The only real root seems to be about 2.2381
On Nov 7, 1:27*pm, kiywrote:
> > 3.x + 5.x^2 + 34.x^3 + 24.x^9 = 34234
>
> > how can i solve like this questions...can anybody help ??
>
> [ 24 0 0 0 0 0 34 5 3 -34234 ] PROOT
>
> The only real root seems to be about 2.2381
More graphical method:
[RS] NUM.SLV(7) 3. Solve poly...
Type in the coefficients as a vector in descending order, then move
the cursor to Roots: and press SOLVE(F6). It returns a vector of the
roots in the screen, then pushes that vector onto level 1 of the stack
and labels it "Roots:".
Use the command OBJ\-> to separate the vector into its entries (in
this case, so that each root gets its own stack level). The command OBJ
\-> also returns {9.}, meaning that there were 9 objects in that
vector. Now it is quick to see that there are 8 complex roots and 1
real root.
Of course, kiy's method of entering the coefficients as a vector and
using the command PROOT is much quicker if you can remember the name
of the command (which is not that difficult). You can also use the OBJ
\-> command on the result from PROOT (which is equivalent to the
NUM.SLV result).
As an aside, a related command is PEVAL. It takes two arguments: the
vector of coefficients in descending order in level 2, and the point
where it is to be evaluated in level 1:
2: [4 0 0 7 2]
1: 5
PEVAL
is shorthand for evaluating 4x^4 + 7x + 2 at x = 5.
S.C.