If α, β, γ are the roots of the given equation then what is the value of ∑α(β+ γ)?

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2 Answers
George C.
Oct 18, 2017

#sumalpha^2beta = 3r-pq#

#sumalpha(beta+gamma) = 2q#

Explanation:

First note that:

#x^3+px^2+qx+r#

#= (x-alpha)(x-beta)(x-gamma)#

#= x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma#

So equating coefficients:

#{ (alpha+beta+gamma = -p), (alphabeta+betagamma+gammaalpha = q), (alphabetagamma = -r) :}#

If I understand correctly, #sumalpha^2beta# is shorthand for:

#alpha^2beta+beta^2gamma+gamma^2alpha+alphabeta^2+betagamma^2+gammaalpha^2#

In which case, note that:

#alpha^2beta+beta^2gamma+gamma^2alpha+alphabeta^2+betagamma^2+gammaalpha^2#

#= (alphabeta+betagamma+gammaalpha)(alpha+beta+gamma)-3alphabetagamma#

#= (q)(-p)-3(-r)#

#= 3r-pq#

Also, #sumalpha(beta+gamma)# is short for:

#alpha(beta+gamma)+beta(gamma+alpha)+gamma(alpha+beta)=2(alphabeta+betagamma+gammaalpha)=2q#

Shwetank Mauria
Oct 18, 2017

#sumalpha(beta+gamma)=2q# and #sumalpha^2beta=3r-pq# i.e. (2)

Explanation:

If #alpha,beta,gamma# are roots of #x^3+px^2+qx+r=0#

then #p=-(alpha+beta+gamma)#, #q=alphabeta+betagamma+gammaalpha# and #r=-alphabetagamma#

One check this by expanding LHS of #(x-alpha)(x-beta)(x-gamma)=x^3+px^2+qx+r# and comparing coefficients of powers of #x#

Hence, #sumalpha(beta+gamma)=sum(alphabeta+alphagamma)#

= #alphabeta+alphagamma+betagamma+betaalpha+gammaalpha+gammabeta#

= #2(alphabeta+betagamma+gammaalpha)=2q#

and for#sumalpha^2beta=alpha^2beta+alpha^2gamma+beta^2gamma+beta^2alpha+gamma^2alpha+gamma^2alpha#

observe that #pq=-(alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)#

= #-(alpha^2beta+alphabetagamma+gammaalpha^2+alphabeta^2+beta^2gamma+alphabetagamma+alphabetagamma+betagamma^2+gamma^2alpha)#

= #-(sumalpha^2beta-3r)#

or #sumalpha^2beta=3r-pq#

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