Question-65289

Question Number 65289 by naka3546 last updated on 27/Jul/19

Commented by Prithwish sen last updated on 27/Jul/19

let 𝚺pqr=A,𝚺pq=B,𝚺p=C Then from the last three equation we get A+B+C=5 A+2B+4C=−3 A+3B+9C=−21 solving this we get A=3,B=7,C=−5 ∴(p+4)(q+4)(r+4)(s+4)=4A+16B+64C+254 =12+112−320+254=58 please check.

$$\mathrm{let} \\ $$$$\boldsymbol{\Sigma\mathrm{pqr}}=\boldsymbol{\mathrm{A}},\boldsymbol{\Sigma\mathrm{pq}}=\boldsymbol{\mathrm{B}},\boldsymbol{\Sigma\mathrm{p}}=\boldsymbol{\mathrm{C}} \\ $$$$\mathrm{Then}\:\mathrm{from}\:\mathrm{the}\:\mathrm{last}\:\mathrm{three}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{5} \\ $$$$\mathrm{A}+\mathrm{2B}+\mathrm{4C}=−\mathrm{3} \\ $$$$\mathrm{A}+\mathrm{3B}+\mathrm{9C}=−\mathrm{21} \\ $$$$\mathrm{solving}\:\mathrm{this}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{A}=\mathrm{3},\mathrm{B}=\mathrm{7},\mathrm{C}=−\mathrm{5} \\ $$$$\therefore\left(\boldsymbol{\mathrm{p}}+\mathrm{4}\right)\left(\boldsymbol{\mathrm{q}}+\mathrm{4}\right)\left(\boldsymbol{\mathrm{r}}+\mathrm{4}\right)\left(\boldsymbol{\mathrm{s}}+\mathrm{4}\right)=\mathrm{4}\boldsymbol{\mathrm{A}}+\mathrm{16}\boldsymbol{\mathrm{B}}+\mathrm{64}\boldsymbol{\mathrm{C}}+\mathrm{254} \\ $$$$=\mathrm{12}+\mathrm{112}−\mathrm{320}+\mathrm{254}=\mathrm{58} \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$

Answered by mr W last updated on 28/Jul/19

pqrs+Σpqr+Σpq+Σp+1=4 ⇒Σpqr+Σpq+Σp=5 ...(i) pqrs+2Σpqr+4Σpq+8Σp+16=8 ⇒Σpqr+2Σpq+4Σp=−3 ...(ii) pqrs+3Σpqr+9Σpq+27Σp+81=16 ⇒Σpqr+3Σpq+9Σp=−21 ...(iii) ⇒Σpqr=3 ⇒Σpq=7 ⇒Σp=−5 (p+4)(q+4)(r+4)(s+4) =pqrs+4Σpqr+16Σpq+64Σp+256 =−2+4×3+16×7+64×(−5)+256 =−2+12+112−320+256 =58 generally (p+n)(q+n)(r+n)(s+n) =pqrs+nΣpqr+n^2 Σpq+n^3 Σp+n^4 =−2+3n+7n^2 −5n^3 +n^4 ============== p,q,r,s are roots of eqn. z^4 +5z^3 +7z^2 −3z−2=0

$${pqrs}+\Sigma{pqr}+\Sigma{pq}+\Sigma{p}+\mathrm{1}=\mathrm{4} \\ $$$$\Rightarrow\Sigma{pqr}+\Sigma{pq}+\Sigma{p}=\mathrm{5}\:\:\:…\left({i}\right) \\ $$$${pqrs}+\mathrm{2}\Sigma{pqr}+\mathrm{4}\Sigma{pq}+\mathrm{8}\Sigma{p}+\mathrm{16}=\mathrm{8} \\ $$$$\Rightarrow\Sigma{pqr}+\mathrm{2}\Sigma{pq}+\mathrm{4}\Sigma{p}=−\mathrm{3}\:\:\:…\left({ii}\right) \\ $$$${pqrs}+\mathrm{3}\Sigma{pqr}+\mathrm{9}\Sigma{pq}+\mathrm{27}\Sigma{p}+\mathrm{81}=\mathrm{16} \\ $$$$\Rightarrow\Sigma{pqr}+\mathrm{3}\Sigma{pq}+\mathrm{9}\Sigma{p}=−\mathrm{21}\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\Sigma{pqr}=\mathrm{3} \\ $$$$\Rightarrow\Sigma{pq}=\mathrm{7} \\ $$$$\Rightarrow\Sigma{p}=−\mathrm{5} \\ $$$$ \\ $$$$\left({p}+\mathrm{4}\right)\left({q}+\mathrm{4}\right)\left({r}+\mathrm{4}\right)\left({s}+\mathrm{4}\right) \\ $$$$={pqrs}+\mathrm{4}\Sigma{pqr}+\mathrm{16}\Sigma{pq}+\mathrm{64}\Sigma{p}+\mathrm{256} \\ $$$$=−\mathrm{2}+\mathrm{4}×\mathrm{3}+\mathrm{16}×\mathrm{7}+\mathrm{64}×\left(−\mathrm{5}\right)+\mathrm{256} \\ $$$$=−\mathrm{2}+\mathrm{12}+\mathrm{112}−\mathrm{320}+\mathrm{256} \\ $$$$=\mathrm{58} \\ $$$$ \\ $$$${generally} \\ $$$$\left({p}+{n}\right)\left({q}+{n}\right)\left({r}+{n}\right)\left({s}+{n}\right) \\ $$$$={pqrs}+{n}\Sigma{pqr}+{n}^{\mathrm{2}} \Sigma{pq}+{n}^{\mathrm{3}} \Sigma{p}+{n}^{\mathrm{4}} \\ $$$$=−\mathrm{2}+\mathrm{3}{n}+\mathrm{7}{n}^{\mathrm{2}} −\mathrm{5}{n}^{\mathrm{3}} +{n}^{\mathrm{4}} \\ $$$$ \\ $$$$============== \\ $$$${p},{q},{r},{s}\:{are}\:{roots}\:{of}\:{eqn}. \\ $$$${z}^{\mathrm{4}} +\mathrm{5}{z}^{\mathrm{3}} +\mathrm{7}{z}^{\mathrm{2}} −\mathrm{3}{z}−\mathrm{2}=\mathrm{0} \\ $$

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