Question-138396

Question Number 138396 by SLVR last updated on 13/Apr/21

Commented by SLVR last updated on 13/Apr/21

$${mr}.{W}..\:{Good}\:{morning} \\ $$$${previous}\:{this}\:{was}\:{given}..{but} \\ $$$${i}\:{could}\:{not}\:{get}\:{it} \\ $$$${kindly}\:{help}\:{me}\:{the}\:{solution} \\ $$

Commented by EDWIN88 last updated on 14/Apr/21

$${q}\mathrm{135556} \\ $$

Answered by MJS_new last updated on 13/Apr/21

(A/(s−p))+(B/(s−q))+(C/(s−r))=(1/((s−p)(s−q)(s−r))) ⇒ A=(1/((q−p)(r−p)))∧B=(1/((p−q)(r−q)))∧C=(1/((p−r)(q−r))) (1/A)+(1/B)+(1/C)=p^2 +q^2 +r^2 −(pq+pr+qr) we know (x−p)(x−q)(x−r)=x^3 −22x^2 +80x−67 ⇒ p+q+r=22 pq+pr+qr=80 pqr=67 (p+q+r)^2 =p^2 +q^2 +r^2 +2(pq+pr+qr)=484 484−3(pq+pr+qr)=484=3×80=244 (⇒ for x^3 +αx^2 +βx+γ=0 the sum (1/A)+(1/B)+(1/C)=α^2 −3β)

$$\frac{{A}}{{s}−{p}}+\frac{{B}}{{s}−{q}}+\frac{{C}}{{s}−{r}}=\frac{\mathrm{1}}{\left({s}−{p}\right)\left({s}−{q}\right)\left({s}−{r}\right)} \\ $$$$\Rightarrow \\ $$$${A}=\frac{\mathrm{1}}{\left({q}−{p}\right)\left({r}−{p}\right)}\wedge{B}=\frac{\mathrm{1}}{\left({p}−{q}\right)\left({r}−{q}\right)}\wedge{C}=\frac{\mathrm{1}}{\left({p}−{r}\right)\left({q}−{r}\right)} \\ $$$$\frac{\mathrm{1}}{{A}}+\frac{\mathrm{1}}{{B}}+\frac{\mathrm{1}}{{C}}={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\left({pq}+{pr}+{qr}\right) \\ $$$$\mathrm{we}\:\mathrm{know} \\ $$$$\left({x}−{p}\right)\left({x}−{q}\right)\left({x}−{r}\right)={x}^{\mathrm{3}} −\mathrm{22}{x}^{\mathrm{2}} +\mathrm{80}{x}−\mathrm{67} \\ $$$$\Rightarrow \\ $$$${p}+{q}+{r}=\mathrm{22} \\ $$$${pq}+{pr}+{qr}=\mathrm{80} \\ $$$${pqr}=\mathrm{67} \\ $$$$\left({p}+{q}+{r}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}\left({pq}+{pr}+{qr}\right)=\mathrm{484} \\ $$$$\mathrm{484}−\mathrm{3}\left({pq}+{pr}+{qr}\right)=\mathrm{484}=\mathrm{3}×\mathrm{80}=\mathrm{244} \\ $$$$ \\ $$$$\left(\Rightarrow\:\mathrm{for}\:{x}^{\mathrm{3}} +\alpha{x}^{\mathrm{2}} +\beta{x}+\gamma=\mathrm{0}\:\mathrm{the}\:\mathrm{sum}\:\frac{\mathrm{1}}{{A}}+\frac{\mathrm{1}}{{B}}+\frac{\mathrm{1}}{{C}}=\alpha^{\mathrm{2}} −\mathrm{3}\beta\right) \\ $$

Commented by SLVR last updated on 13/Apr/21

$${Thanks}\:{mr}.{MJS}.{NEW} \\ $$

Commented by SLVR last updated on 13/Apr/21

$${and}\:{mr}.{W}.{So}\:{kind}\:{of}\:{you} \\ $$

Answered by mr W last updated on 13/Apr/21

p+q+r=22 pq+qr+rp=80 pqr=67 (p+q+r)^2 =p^2 +q^2 +r^2 +2(pq+qr+rp) ⇒p^2 +q^2 +r^2 =(p+q+r)^2 −2(pq+qr+rp) (1/((s−p)(s−q)(s−r)))=(A/(s−p))+(B/(s−q))+(C/(s−r)) =((A(s−q)(s−r)+B(s−p)(s−r)+C(s−p)(s−q))/((s−p)(s−q)(s−r))) A(s−q)(s−r)+B(s−p)(s−r)+C(s−p)(s−q)=1 s=p ⇒A=(1/((p−q)(p−r)))⇒(1/A)=(p−q)(p−r)=p^2 −pq−rp+qr s=q ⇒B=(1/((q−p)(q−r)))⇒(1/B)=(q−p)(q−r)=q^2 −pq−qr+rp s=r ⇒C=(1/((r−p)(r−q)))⇒(1/C)=(r−p)(r−q)=r^2 −qr−rp+pq (1/A)+(1/B)+(1/C)=p^2 +q^2 +r^2 −(pq+qr+rp) =(p+q+r)^2 −3(pq+qr+rp) =22^2 −3×80=244 ⇒answer b)

$${p}+{q}+{r}=\mathrm{22} \\ $$$${pq}+{qr}+{rp}=\mathrm{80} \\ $$$${pqr}=\mathrm{67} \\ $$$$\left({p}+{q}+{r}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}\left({pq}+{qr}+{rp}\right) \\ $$$$\Rightarrow{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\left({p}+{q}+{r}\right)^{\mathrm{2}} −\mathrm{2}\left({pq}+{qr}+{rp}\right) \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\left({s}−{p}\right)\left({s}−{q}\right)\left({s}−{r}\right)}=\frac{{A}}{{s}−{p}}+\frac{{B}}{{s}−{q}}+\frac{{C}}{{s}−{r}} \\ $$$$=\frac{{A}\left({s}−{q}\right)\left({s}−{r}\right)+{B}\left({s}−{p}\right)\left({s}−{r}\right)+{C}\left({s}−{p}\right)\left({s}−{q}\right)}{\left({s}−{p}\right)\left({s}−{q}\right)\left({s}−{r}\right)} \\ $$$${A}\left({s}−{q}\right)\left({s}−{r}\right)+{B}\left({s}−{p}\right)\left({s}−{r}\right)+{C}\left({s}−{p}\right)\left({s}−{q}\right)=\mathrm{1} \\ $$$${s}={p}\:\Rightarrow{A}=\frac{\mathrm{1}}{\left({p}−{q}\right)\left({p}−{r}\right)}\Rightarrow\frac{\mathrm{1}}{{A}}=\left({p}−{q}\right)\left({p}−{r}\right)={p}^{\mathrm{2}} −{pq}−{rp}+{qr} \\ $$$${s}={q}\:\Rightarrow{B}=\frac{\mathrm{1}}{\left({q}−{p}\right)\left({q}−{r}\right)}\Rightarrow\frac{\mathrm{1}}{{B}}=\left({q}−{p}\right)\left({q}−{r}\right)={q}^{\mathrm{2}} −{pq}−{qr}+{rp} \\ $$$${s}={r}\:\Rightarrow{C}=\frac{\mathrm{1}}{\left({r}−{p}\right)\left({r}−{q}\right)}\Rightarrow\frac{\mathrm{1}}{{C}}=\left({r}−{p}\right)\left({r}−{q}\right)={r}^{\mathrm{2}} −{qr}−{rp}+{pq} \\ $$$$\frac{\mathrm{1}}{{A}}+\frac{\mathrm{1}}{{B}}+\frac{\mathrm{1}}{{C}}={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\left({pq}+{qr}+{rp}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({p}+{q}+{r}\right)^{\mathrm{2}} −\mathrm{3}\left({pq}+{qr}+{rp}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{22}^{\mathrm{2}} −\mathrm{3}×\mathrm{80}=\mathrm{244} \\ $$$$\left.\Rightarrow{answer}\:{b}\right) \\ $$

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