Question Number 165651 by MathsFan last updated on 06/Feb/22
$${If}\:{the}\:{equation}\:\left(\mathrm{2}{x}−\mathrm{1}\right)−{p}\left({x}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0}, \\ $$$${where}\:{p}\:{is}\:{constant},\:{has}\:{imaginary} \\ $$$${roots},\:{deduce}\:{that}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}\geqslant\mathrm{0}. \\ $$
Commented by mr W last updated on 06/Feb/22
$${px}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}{p}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\pm\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{4}{p}\left(\mathrm{2}{p}+\mathrm{1}\right)}}{\mathrm{2}{p}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−{p}−\mathrm{2}{p}^{\mathrm{2}} }}{{p}} \\ $$$${if}\:\mathrm{1}−{p}−\mathrm{2}{p}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow{real}\:{roots} \\ $$$${if}\:\mathrm{1}−{p}−\mathrm{2}{p}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow{complex}\:{roots}: \\ $$$$\:\:\:\:{x}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}}}{{p}} \\ $$$$\:\:\:\:{since}\:\frac{\mathrm{1}}{{p}}\neq\mathrm{0},\:{the}\:{equation}\:{can}\:{never} \\ $$$$\:\:\:\:{have}\:{imaginary}\:{roots}! \\ $$$$ \\ $$$${therefore}:\:{question}\:{is}\:{wrong}! \\ $$$$\left.\mathrm{1}\right)\:{when}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}=\mathrm{0}\:{it}\:{has}\:{real}\:{roots}. \\ $$$$\left.\mathrm{2}\right)\:{when}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}>\mathrm{0}\:{it}\:{has}\:{comlex}\:{roots}. \\ $$
Commented by peter frank last updated on 06/Feb/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by MathsFan last updated on 06/Feb/22
$${thanks} \\ $$
Commented by chrisbridge last updated on 06/Feb/22
$${i}\:{think}\:{you}\:{do}\:{not}\:{need}\:\frac{\mathrm{1}}{{p}}\:{to}\:{be}\:{equal}\:{to}\:\mathrm{0}\:{to}\:{have}\:{imaginary}\:{roots}.\:{You}\:{only}\:{need}\:{the}\:{discrimant}\:{to}\:{be}\:=\:\mathrm{0}.\:{From}\:{your}\:{work}\:{above}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}>\mathrm{0}\:{is}\:{true}\:{for}\:{the}\:{first}\:{equation}\:{to}\:{have}\:{imaginary}\:{roots} \\ $$
Commented by mr W last updated on 06/Feb/22
$${i}\:{don}'{t}\:{need},\:{but}\:{the}\:{question}\:{needs}! \\ $$$${the}\:{question}\:{requests}\:{imaginary}\:{roots}. \\ $$$${bi}\:{is}\:{an}\:{imaginary}\:{number}. \\ $$$${a}+{bi}\:{is}\:{a}\:{complex}\:{number},\:{not}\:{an} \\ $$$${imaginary}\:{number},\:{if}\:{a}\neq\mathrm{0}. \\ $$$${since}\:\frac{\mathrm{1}}{{p}}\:{can}'{t}\:{be}\:{zero},\:{so}\:{the}\:{equation} \\ $$$${can}\:{never}\:{have}\:{imaginary}\:{roots}! \\ $$
Answered by chrisbridge last updated on 06/Feb/22
$${b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0}\:{for}\:{imaginary}\:{roots} \\ $$$$\mathrm{2}{x}\:−\:\mathrm{1}\:−{px}^{\mathrm{2}} \:−\mathrm{2}{p}\:=\:\mathrm{0} \\ $$$${px}^{\mathrm{2}} −\mathrm{2}{x}+\left(\mathrm{1}+\mathrm{2}{p}\right)=\mathrm{0} \\ $$$$\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}{p}\left(\mathrm{1}+\mathrm{2}{p}\right)<\mathrm{0} \\ $$$$\mathrm{4}−\mathrm{4}{p}−\mathrm{8}{p}^{\mathrm{2}} <\mathrm{0} \\ $$$$\mathrm{4}−\mathrm{4}{p}−\mathrm{8}{p}^{\mathrm{2}} <\mathrm{0} \\ $$$${dividing}\:{through}\:{by}\:\mathrm{4} \\ $$$$\mathrm{1}−{p}−\mathrm{2}{p}^{\mathrm{2}} <\mathrm{0} \\ $$$${on}\:{multiplying}\:{through}\:{by}\:\left(−\mathrm{1}\right) \\ $$$$−\mathrm{1}+{p}+\mathrm{2}{p}^{\mathrm{2}} >\mathrm{0} \\ $$$${rearraging} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}>\mathrm{0} \\ $$$${required}\:{solution} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 07/Feb/22
$${you}\:{seem}\:{to}\:{think}\:\:{that}\:{imaginary}\: \\ $$$${numbers}\:{and}\:{complex}\:{numbers}\:{are}\: \\ $$$${the}\:{same}\:{thing}. \\ $$