Question Number 27061 by math solver last updated on 01/Jan/18
Commented by math solver last updated on 02/Jan/18
$${pLz}\:{help}? \\ $$
Answered by prakash jain last updated on 02/Jan/18
$${x}^{\mathrm{2}} +{px}+{q}^{\mathrm{2}} =\mathrm{0} \\ $$$${root}\:{u},{v}\:\:\:{u},{v}\in\mathbb{C} \\ $$$${u}\bar {{u}}={v}\bar {{v}}\Rightarrow{u}=\frac{{v}\bar {{v}}}{\mathrm{2}}\:\:\left({A}\right) \\ $$$${u}+{v}=−{p}\Rightarrow{p}^{\mathrm{2}} ={u}^{\mathrm{2}} +{v}^{\mathrm{2}} +\mathrm{2}{uv} \\ $$$${uv}={q}^{\mathrm{2}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\frac{{u}}{{v}}+\frac{{v}}{{u}}+\mathrm{2}\:\:\left({B}\right) \\ $$$${subtitute}\:{u}\:{from}\:\left({A}\right)\:{to}\:\left({B}\right) \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\frac{{v}\bar {{v}}}{{v}\bar {{u}}}+\frac{{v}\bar {{u}}}{{v}\bar {{v}}}+\mathrm{2} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\frac{\bar {{v}}}{\bar {{u}}}+\frac{\bar {{u}}}{\bar {{v}}}+\mathrm{2}\:\:\:\:\:\left({C}\right) \\ $$$$\left({B}\right)+\left({C}\right) \\ $$$$\mathrm{2}\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\left(\frac{{u}}{{v}}+\frac{\bar {{u}}}{\bar {{v}}}\right)+\left(\frac{{v}}{{u}}+\frac{\bar {{v}}}{\bar {{u}}}\right)+\mathrm{4} \\ $$$$\mathrm{2}\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\Re\left(\frac{{u}}{{v}}\right)+\Re\left(\frac{{v}}{{u}}\right)+\mathrm{4}=\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{will}\:\mathrm{need}\:\mathrm{to}\:\mathrm{prove}\:\Re\left(\frac{{u}}{{v}}\right)+\Re\left(\frac{{v}}{{u}}\right)>−\mathrm{4} \\ $$
Answered by shiv15031973@gmail.com last updated on 05/Jan/18
$${rational}\:{no}.\:{as}\:{it}\:{is}\:{written}\:{in}\:{the} \\ $$$$\:{form}\:{of}\frac{{p}}{{q}} \\ $$
Commented by prakash jain last updated on 05/Jan/18
$${p}\:\mathrm{and}\:{q}\:\mathrm{are}\:\mathrm{given}\:\mathrm{as}\:\mathrm{complex}. \\ $$