impossible polynomial (1 Viewer)

MC Squidge · May 16, 2009

solve x^3-5x^2+11=0

Timothy.Siu · May 16, 2009

how is x=2 a zero
edit: lol he deleted his post

shaon0 · May 16, 2009

Timothy.Siu said:
how is x=2 a zero
edit: lol he deleted his post

yeah sorry about that. It's been a while since polynomial, For factor theorem to work, x=(+-11,+-1) should work since 11 is prime but it doesn't. I think it'll have some really big irrational solution if the roots are real.

Timothy.Siu · May 16, 2009

yeah thats right.
so the original question is dumb basically.

shaon0 · May 16, 2009

Timothy.Siu said:
yeah thats right.
so the original question is dumb basically.

Probably have to approximate the roots of the equation.

MC Squidge · May 17, 2009

so the roots can be irrational?

shaon0 · May 17, 2009

MC Squidge said:
so the roots can be irrational?

Yeah, since the degree of the polynomial is 3 and the co-efficients are real. There can be unreal roots.

cutemouse · May 17, 2009

Here, I've done all the hard work....

$x=\frac{5}{3}+\frac{25}{3 \left(\frac{1}{2} \left(-47+3 i \sqrt{6699}\right)\right)^{1/3}}+\frac{1}{3} \left(\frac{1}{2} \left(-47+3 i \sqrt{6699}\right)\right)^{1/3}\\x=\frac{5}{3}-\frac{1}{6} \left(1+i \sqrt{3}\right) \left(\frac{1}{2} \left(-47+3 i \sqrt{6699}\right)\right)^{1/3}-\frac{25 \left(1-i \sqrt{3}\right)}{3 2^{2/3} \left(-47+3 i \sqrt{6699}\right)^{1/3}}\\x=\frac{5}{3}-\frac{1}{6} \left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(-47+3 i \sqrt{6699}\right)\right)^{1/3}-\frac{25 \left(1+i \sqrt{3}\right)}{3 2^{2/3} \left(-47+3 i \sqrt{6699}\right)^{1/3}}$

So, it's not impossible... All you need is a bit of persistance and patience =D (and some other stuff =P)

shaon0 · May 17, 2009

jm01 said:
Here, I've done all the hard work....

$x=\frac{5}{3}+\frac{25}{3 \left(\frac{1}{2} \left(-47+3 i \sqrt{6699}\right)\right)^{1/3}}+\frac{1}{3} \left(\frac{1}{2} \left(-47+3 i \sqrt{6699}\right)\right)^{1/3}\\x=\frac{5}{3}-\frac{1}{6} \left(1+i \sqrt{3}\right) \left(\frac{1}{2} \left(-47+3 i \sqrt{6699}\right)\right)^{1/3}-\frac{25 \left(1-i \sqrt{3}\right)}{3 2^{2/3} \left(-47+3 i \sqrt{6699}\right)^{1/3}}\\x=\frac{5}{3}-\frac{1}{6} \left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(-47+3 i \sqrt{6699}\right)\right)^{1/3}-\frac{25 \left(1+i \sqrt{3}\right)}{3 2^{2/3} \left(-47+3 i \sqrt{6699}\right)^{1/3}}$

So, it's not impossible... All you need is a bit of persistance and patience =D (and some other stuff =P)

looks like 'you' used the cubic formula.

cutemouse · May 17, 2009

Nope, I didn't =D

I used a much easier method/way

riseek · May 17, 2009

could u please explain

jm01 said:
Nope, I didn't =D

I used a much easier method/way

jet · May 17, 2009

he.... made someone else use the cubic formula and he typed out the answers????

tommykins · May 17, 2009

x = -1.319350320, 1.876666879, 4.442683441

shaon0 · May 17, 2009

jm01 said:
Nope, I didn't =D

I used a much easier method/way

online polynomial solver?

cutemouse · May 17, 2009

shaon0 said:
online polynomial solver?

Close - I have Wolfram Mathematica on my computer =D Nice thing to have, once you get to know how to use it. =P

omniscience · May 23, 2009

your face is impossible bro

Dragonmaster262 · May 23, 2009

MC Squidge said:
solve x^3-5x^2+11=0

Did you get this from Cambridge?

MC Squidge · May 23, 2009

no i got it from a friend who asked me it and laughed at me when i couldnt solve it

impossible polynomial (1 Viewer)

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