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Find-the-relation-between-p-q-and-r-if-one-of-the-root-of-the-equation-px-2-qx-r-0-is-double-the-other-




Question Number 52573 by Necxx last updated on 09/Jan/19
Find the relation between p q and r  if one of the root of the equation  px^2 +qx+r=0 is double the other.
$${Find}\:{the}\:{relation}\:{between}\:{p}\:{q}\:{and}\:{r} \\ $$$${if}\:{one}\:{of}\:{the}\:{root}\:{of}\:{the}\:{equation} \\ $$$${px}^{\mathrm{2}} +{qx}+{r}=\mathrm{0}\:{is}\:{double}\:{the}\:{other}. \\ $$
Answered by mr W last updated on 09/Jan/19
px^2 +qx+r=p(x−a)(x−2a)  px^2 +qx+r=p(x^2 −3ax+2a^2 )  ⇒q=−3pa  ⇒r=2pa^2   ⇒r=2p(−(q/(3p)))^2 =((2q^2 )/(9p))  ⇒9pr=2q^2
$${px}^{\mathrm{2}} +{qx}+{r}={p}\left({x}−{a}\right)\left({x}−\mathrm{2}{a}\right) \\ $$$${px}^{\mathrm{2}} +{qx}+{r}={p}\left({x}^{\mathrm{2}} −\mathrm{3}{ax}+\mathrm{2}{a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{q}=−\mathrm{3}{pa} \\ $$$$\Rightarrow{r}=\mathrm{2}{pa}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{2}{p}\left(−\frac{{q}}{\mathrm{3}{p}}\right)^{\mathrm{2}} =\frac{\mathrm{2}{q}^{\mathrm{2}} }{\mathrm{9}{p}} \\ $$$$\Rightarrow\mathrm{9}{pr}=\mathrm{2}{q}^{\mathrm{2}} \\ $$
Commented by Necxx last updated on 10/Jan/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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