Question Number 158829 by MathsFan last updated on 09/Nov/21
$$\:{The}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\:\mathrm{2}{x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\:{are}\:\mathrm{2}\alpha+\beta\:{and} \\ $$$$\:\alpha+\mathrm{2}\beta.\:{Calculate}\:{the}\:{values}\:{of} \\ $$$$\:{p}\:{and}\:{q} \\ $$
Commented by Rasheed.Sindhi last updated on 09/Nov/21
$${Numerical}\:{answers}\:{of}\:{p}\:\&\:{q}\:{are}\:{not} \\ $$$${possible}\:{because}\:{data}\:{is}\:{insufficient}. \\ $$
Commented by MathsFan last updated on 09/Nov/21
$$\mathrm{yeah},\:\mathrm{sure} \\ $$
Answered by ajfour last updated on 09/Nov/21
$$\left(\mathrm{2}\alpha+\beta\right)+\left(\alpha+\mathrm{2}\beta\right)=\mathrm{3}\left(\alpha+\beta\right)=−\frac{{p}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\alpha+\beta\right)\left(\alpha+\mathrm{2}\beta\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left(\alpha+\beta\right)^{\mathrm{2}} +\alpha\beta=\frac{{q}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\frac{{p}^{\mathrm{2}} }{\mathrm{18}}+\alpha\beta=\frac{{q}}{\mathrm{2}} \\ $$$${p}=−\mathrm{6}\left(\alpha+\beta\right)\:\:\:;\:{q}=\frac{{p}^{\mathrm{2}} }{\mathrm{9}}+\mathrm{2}\alpha\beta \\ $$
Commented by MathsFan last updated on 09/Nov/21
$$\mathrm{thank}\:\mathrm{you} \\ $$