Question Number 228126 by fantastic2 last updated on 28/Mar/26
$${if}\:\:\alpha\:\&\:\beta\:{are}\:{two}\:{roots}\:{of}\:{equation} \\ $$$${x}^{\mathrm{2}} +{px}+\mathrm{1}=\mathrm{0} \\ $$$${and}\:\gamma\:\&\:\delta\:{are}\:{two}\:{roots}\:{of}\:{equation} \\ $$$${x}^{\mathrm{2}} +{qx}+\mathrm{1}=\mathrm{0} \\ $$$$\:{then}\:{prove}\:{that}\: \\ $$$$\left(\alpha−\gamma\right)\left(\beta−\gamma\right)\left(\alpha+\delta\right)\left(\beta+\delta\right)={q}^{\mathrm{2}} −{p}^{\mathrm{2}} \\ $$
Answered by TonyCWX last updated on 29/Mar/26
![α + β = −p αβ = 1 γ + δ = −q γδ = 1 (α − γ)(β − γ)(α + δ)(β + δ) = (αβ − γ(α + β) + γ^2 )(αβ + δ(α + β) + δ^2 ) = (γ^2 + pγ + 1)(δ^2 − pδ + 1) = γ^2 δ^2 − γ^2 δp + γ^2 + pγδ^2 − p^2 γδ + pγ + δ^2 − pδ + 1 = 1 − γp + γ^2 + pδ − p^2 + pγ + δ^2 − pδ + 1 = 2 − p^2 + [(γ + δ)^2 − 2γδ] = 2 − p^2 + q^2 − 2 = q^2 − p^2 [Q.E.D]](https://www.tinkutara.com/question/Q228127.png)
$$\alpha\:+\:\beta\:=\:−{p} \\ $$$$\alpha\beta\:=\:\mathrm{1} \\ $$$$\gamma\:+\:\delta\:=\:−{q} \\ $$$$\gamma\delta\:=\:\mathrm{1} \\ $$$$ \\ $$$$\left(\alpha\:−\:\gamma\right)\left(\beta\:−\:\gamma\right)\left(\alpha\:+\:\delta\right)\left(\beta\:+\:\delta\right) \\ $$$$=\:\left(\alpha\beta\:−\:\gamma\left(\alpha\:+\:\beta\right)\:+\:\gamma^{\mathrm{2}} \right)\left(\alpha\beta\:+\:\delta\left(\alpha\:+\:\beta\right)\:+\:\delta^{\mathrm{2}} \right) \\ $$$$=\:\left(\gamma^{\mathrm{2}} \:+\:{p}\gamma\:+\:\mathrm{1}\right)\left(\delta^{\mathrm{2}} \:−\:{p}\delta\:+\:\mathrm{1}\right) \\ $$$$=\:\gamma^{\mathrm{2}} \delta^{\mathrm{2}} \:−\:\gamma^{\mathrm{2}} \delta{p}\:+\:\gamma^{\mathrm{2}} \:+\:{p}\gamma\delta^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \gamma\delta\:+\:{p}\gamma\:+\:\delta^{\mathrm{2}} \:−\:{p}\delta\:+\:\mathrm{1} \\ $$$$=\:\mathrm{1}\:−\:\gamma{p}\:+\:\gamma^{\mathrm{2}} \:+\:{p}\delta\:−\:{p}^{\mathrm{2}} \:+\:{p}\gamma\:+\:\delta^{\mathrm{2}} \:−\:{p}\delta\:+\:\mathrm{1} \\ $$$$=\:\mathrm{2}\:−\:{p}^{\mathrm{2}} \:+\:\left[\left(\gamma\:+\:\delta\right)^{\mathrm{2}} \:−\:\mathrm{2}\gamma\delta\right] \\ $$$$=\:\mathrm{2}\:−\:{p}^{\mathrm{2}} \:+\:{q}^{\mathrm{2}} \:−\:\mathrm{2} \\ $$$$=\:{q}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \:\left[{Q}.{E}.{D}\right] \\ $$