Question Number 106138 by bobhans last updated on 03/Aug/20
$$\mathrm{If}\:\mathrm{root}\:\mathrm{of}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{3}} −\mathrm{px}^{\mathrm{2}} +\mathrm{qx}−\mathrm{r}=\mathrm{0}\:\mathrm{are}\:\mathrm{in} \\ $$$$\mathrm{AP}\:\mathrm{than}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{p},\mathrm{q} \\ $$$$\mathrm{and}\:\mathrm{r}\:? \\ $$
Answered by bemath last updated on 03/Aug/20
![let x_1 −d, x_1 ; x_1 +d is the roots of the given equation . we have ⇒(x_1 −d)+x_1 +(x_1 +d) = p [ Vietha′s rule] ⇒3x_1 = p →x_1 = (p/3), so we get ((p/3))^3 −p((p/3))^2 +q((p/3))−r=0 (p^3 /(27))−(p^3 /9)+((pq)/3)−r=0 −2p^3 +9pq−27r=0 27p^3 −9pq+27r=0](https://www.tinkutara.com/question/Q106140.png)
$$\mathrm{let}\:\mathrm{x}_{\mathrm{1}} −\mathrm{d},\:\mathrm{x}_{\mathrm{1}} \:;\:\mathrm{x}_{\mathrm{1}} +\mathrm{d}\:\mathrm{is}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:.\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\Rightarrow\left(\mathrm{x}_{\mathrm{1}} −\mathrm{d}\right)+\mathrm{x}_{\mathrm{1}} +\left(\mathrm{x}_{\mathrm{1}} +\mathrm{d}\right)\:=\:\mathrm{p}\:\left[\:\mathrm{Vietha}'\mathrm{s}\:\mathrm{rule}\right] \\ $$$$\Rightarrow\mathrm{3x}_{\mathrm{1}} \:=\:\mathrm{p}\:\rightarrow\mathrm{x}_{\mathrm{1}} \:=\:\frac{\mathrm{p}}{\mathrm{3}},\:\mathrm{so}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\left(\frac{\mathrm{p}}{\mathrm{3}}\right)^{\mathrm{3}} −\mathrm{p}\left(\frac{\mathrm{p}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{q}\left(\frac{\mathrm{p}}{\mathrm{3}}\right)−\mathrm{r}=\mathrm{0} \\ $$$$\frac{\mathrm{p}^{\mathrm{3}} }{\mathrm{27}}−\frac{\mathrm{p}^{\mathrm{3}} }{\mathrm{9}}+\frac{\mathrm{pq}}{\mathrm{3}}−\mathrm{r}=\mathrm{0} \\ $$$$−\mathrm{2p}^{\mathrm{3}} +\mathrm{9pq}−\mathrm{27r}=\mathrm{0} \\ $$$$\mathrm{27p}^{\mathrm{3}} −\mathrm{9pq}+\mathrm{27r}=\mathrm{0} \\ $$
Commented by bemath last updated on 03/Aug/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 03/Aug/20
$${Why}\:{have}\:{you}\:{assumed}\:{one}\:{root} \\ $$$$\:{r}\:\left({the}\:{contant}\:{term}\:{of}\:{the}\right. \\ $$$$\left.{equation}\right)? \\ $$
Commented by Rasheed.Sindhi last updated on 03/Aug/20
$$\mathcal{N}{ice}! \\ $$
Commented by bemath last updated on 03/Aug/20
$$\mathrm{oo}\:\mathrm{sorry}.\:\mathrm{i}\:\mathrm{can}\:\mathrm{edit}\: \\ $$