Re: International Space Station

Does anyone enjoy algebra here..

If so..what is the answer to this equation?

2x^3-3x^2+px+q = 0 substitute 3 for x 54-27+3p+q = 0 substitute -2 for x -16-12-2p+q = 0 27+3p+q = 0 -28-2p+q = 0 55+5p = 0 p=-11 q=6??

Re: International Space Station

Blimey Lilac, I don't know how you wrote all that let alone understood it!

Re: International Space Station

That is great. Trouble is when I viewed it it showed blue. Can't wait for the real thing .

Re: International Space Station

http://www.n2yo.com/space-station/

12:45 hrs 16th May.

An internal camera on the ISS is active in the area where the EVA suits are stored. This suggest there is going to be an EVA (External Vehicular Activity) either today or tomorrow.

Re: International Space Station

12:45
Video link cut, but audio link still connected.
.... You just have to be a little patient...

Re: International Space Station

Video but silent. Over north Pacific

Re: International Space Station

A chicken tikka vindaloo tastes great on a Friday evening?

Re: International Space Station

http://www.n2yo.com/space-station/

12:12 Wednesday.

ISS presently over South China, and experiment going on.
Not quite sure what it is...

Re: International Space Station

Ok this looks confusing because you wrote it all out in one line. It should look like this:

Equation

2x^3-3x^2+px+q = 0

If x = 3 then it simplifies as 54-27+3p+q = 0

which simplifies again to 27+3p+q = 0


If x = -2 then it simplifies as -16-12-2p+q = 0

which simplifies again to -28+2p+q = 0


Since both equations in bold = 0 they are the same so we can combine and write:

27+3p+q = -28-2p+q

We can now rearrange that to:

55+5p = 0

Thus 5p = -55

p = -11


So now let's put that value of p back into those 2 bolded equations

When x = 3
27+3p+q = 0

So 27-33+q = 0, So q = 33-27

So q = 6


When x = -2
-28+2p+q = 0

So -28+22+q = 0, So q = 28-22

So q = 6

So in effect everything you posted contained the equations and the answers so I'm not sure what question you were actually asking.

You started with 1 equation with 3 unknown variables in it (x,p and q) and you looked at the cases where x=3 and x=-2

This showed that p = -11 in those 2 cases and that q must thus be 6

What further analysis were you looking for?

Re: International Space Station

I guess you can put those values of p and q back into the original equation i.e.

2x^3-3x^2+px+q=0

So

2x^3-3x^2-11x+6=0

This is now a "cubic" equation which can be solved. I just plugged the values into an online solver which produced the 3 values:

X1 = 3
X2 = -2
X3 = 0.5