Poly root problem, bit stumped

[Xibu4]

I know that [math]\gamma + \delta =-\frac{5}{6}[/math] and [math]\alpha+\beta =0[/math]
so then i subbed that into
\[math]alpha\beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta = -4[/math]
and got that
[math]\alpha \beta + \gamma \delta =-4[/math]
When I sub that all in and factor i get:
[math](\alpha +\beta )(\gamma + \delta ) = 0[/math]
which doesn't really tell me anything

 

[Steven G]

Until you define your variables, I will not read your entire post.

 

[Harry_the_cat]

@Xibu4, please check your initial post. There seems to be something cut off at the end.

 

[Harry_the_cat]

Or is that the start of the next question??

 

[Otis]

got that αβ+γδ=−4

Hi Xibu4. Good. You also know that

α + β = 0
and
γ + δ = -5/6

Solve those equations for β and δ, and then substitute the results in

αβ + γδ = −4

So, you've already used the formula for sum of root products taken two at a time, for a polynomial Ax⁴+Bx³+Cx²+Dx+E:

αβ + βγ + γδ + δα + αγ + βδ = C/A

Have you seen the formula for sum of root products taken three at a time?

αβγ + βγδ + γδα + αβδ = -D/A

Again, make the same substitutions for β and δ.

You now ought to have one equation containing α only, and another equation containing α and γ. Solve that system.

[imath]\;[/imath]

 

[Xibu4]

Or is that the start of the next question??

That's the start of the next question, couldn't crop it out unfortunately.

 

[Xibu4]

Until you define your variables, I will not read your entire post.

I don't really understand what you mean by that, isn't the greek alphabet always used to denote roots of a polynomial.

 

[Xibu4]

Hi Xibu4. Good. You also know that

α + β = 0
and
γ + δ = -5/6

Solve those equations for β and δ, and then substitute the results in

αβ + γδ = −4

So, you've already used the formula for sum of root products taken two at a time, for a polynomial Ax⁴+Bx³+Cx²+Dx+E:

αβ + βγ + γδ + δα + αγ + βδ = C/A

Have you seen the formula for sum of root products taken three at a time?

αβγ + βγδ + γδα + αβδ = -D/A

Again, make the same substitutions for β and δ.

You now ought to have one equation containing α only, and another equation containing α and γ. Solve that system.

[imath]\;[/imath]

Ok thanks I'll try that out now

 

[lex]

Equivalently
[imath](x^2+\tfrac{5}{6}x+\gamma\delta)(x^2+\alpha\beta)=0[/imath] has the same 4 roots.
[imath](6x^2+5x+6\gamma\delta)(x^2+\alpha\beta)=0[/imath]
Multiply this out and compare coefficients with the original polynomial and along with [imath]\alpha=-\beta, \text{ and }\gamma=-\tfrac{5}{6}-\delta[/imath]
this should give equations for [imath]\alpha, \beta, \gamma, \delta[/imath] which can easily be solved.