Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 106138 by bobhans last updated on 03/Aug/20

If root of equation x^3 −px^2 +qx−r=0 are in  AP than what is the relation between p,q  and r ?

$$\mathrm{If}\:\mathrm{root}\:\mathrm{of}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{3}} −\mathrm{px}^{\mathrm{2}} +\mathrm{qx}−\mathrm{r}=\mathrm{0}\:\mathrm{are}\:\mathrm{in} \\ $$$$\mathrm{AP}\:\mathrm{than}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{p},\mathrm{q} \\ $$$$\mathrm{and}\:\mathrm{r}\:? \\ $$

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

$$\mathrm{let}\:\mathrm{x}_{\mathrm{1}} −\mathrm{d},\:\mathrm{x}_{\mathrm{1}} \:;\:\mathrm{x}_{\mathrm{1}} +\mathrm{d}\:\mathrm{is}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:.\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\Rightarrow\left(\mathrm{x}_{\mathrm{1}} −\mathrm{d}\right)+\mathrm{x}_{\mathrm{1}} +\left(\mathrm{x}_{\mathrm{1}} +\mathrm{d}\right)\:=\:\mathrm{p}\:\left[\:\mathrm{Vietha}'\mathrm{s}\:\mathrm{rule}\right] \\ $$$$\Rightarrow\mathrm{3x}_{\mathrm{1}} \:=\:\mathrm{p}\:\rightarrow\mathrm{x}_{\mathrm{1}} \:=\:\frac{\mathrm{p}}{\mathrm{3}},\:\mathrm{so}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\left(\frac{\mathrm{p}}{\mathrm{3}}\right)^{\mathrm{3}} −\mathrm{p}\left(\frac{\mathrm{p}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{q}\left(\frac{\mathrm{p}}{\mathrm{3}}\right)−\mathrm{r}=\mathrm{0} \\ $$$$\frac{\mathrm{p}^{\mathrm{3}} }{\mathrm{27}}−\frac{\mathrm{p}^{\mathrm{3}} }{\mathrm{9}}+\frac{\mathrm{pq}}{\mathrm{3}}−\mathrm{r}=\mathrm{0} \\ $$$$−\mathrm{2p}^{\mathrm{3}} +\mathrm{9pq}−\mathrm{27r}=\mathrm{0} \\ $$$$\mathrm{27p}^{\mathrm{3}} −\mathrm{9pq}+\mathrm{27r}=\mathrm{0} \\ $$

Commented by bemath last updated on 03/Aug/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 03/Aug/20

Why have you assumed one root   r (the contant term of the  equation)?

$${Why}\:{have}\:{you}\:{assumed}\:{one}\:{root} \\ $$$$\:{r}\:\left({the}\:{contant}\:{term}\:{of}\:{the}\right. \\ $$$$\left.{equation}\right)? \\ $$

Commented by Rasheed.Sindhi last updated on 03/Aug/20

Nice!

$$\mathcal{N}{ice}! \\ $$

Commented by bemath last updated on 03/Aug/20

oo sorry. i can edit

$$\mathrm{oo}\:\mathrm{sorry}.\:\mathrm{i}\:\mathrm{can}\:\mathrm{edit}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com