Question Number 187980 by Mingma last updated on 24/Feb/23
Answered by Rasheed.Sindhi last updated on 24/Feb/23
$$\bullet{x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\: \\ $$$$\alpha+\beta=−{b}\:,\:\:\:\alpha\beta={c} \\ $$$$\bullet{x}^{\mathrm{2}} +{bx}−{c}=\mathrm{0} \\ $$$$\gamma+\delta=−{b}\:\:,\:\:\gamma\delta=−{c} \\ $$$$\alpha+\beta=\gamma+\delta\:\:,\:\alpha\beta=−\gamma\delta \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\mathrm{2}\alpha\beta \\ $$$$\:\:\:\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} ={b}^{\mathrm{2}} −\mathrm{2}{c} \\ $$$$\left(\gamma+\delta\right)^{\mathrm{2}} =\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} +\mathrm{2}\gamma\delta \\ $$$$\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} ={b}^{\mathrm{2}} +\mathrm{2}{c} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta+\left(\gamma+\delta\right)^{\mathrm{2}} −\mathrm{2}\gamma\delta=\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\cancel{\mathrm{2}\left({c}\right)}+\left(\gamma+\delta\right)^{\mathrm{2}} −\cancel{\mathrm{2}\left(−{c}\right)}=\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} +\left(\gamma+\delta\right)^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$
Commented by Mingma last updated on 24/Feb/23
Well explained, sir!
Commented by Mingma last updated on 24/Feb/23
Can you show the last part?
Commented by Rasheed.Sindhi last updated on 24/Feb/23
$$\left.\mathcal{T}{his}\:{is}\:{just}\:\mathrm{2}{b}^{\mathrm{2}} ={b}^{\mathrm{2}} +{b}^{\mathrm{2}} \::\right) \\ $$