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Question-187980




Question Number 187980 by Mingma last updated on 24/Feb/23
Answered by Rasheed.Sindhi last updated on 24/Feb/23
•x^2 +bx+c=0   α+β=−b ,   αβ=c  •x^2 +bx−c=0  γ+δ=−b  ,  γδ=−c  α+β=γ+δ  , αβ=−γδ  (α+β)^2 =α^2 +β^2 +2αβ      α^2 +β^2 =b^2 −2c  (γ+δ)^2 =γ^2 +δ^2 +2γδ  γ^2 +δ^2 =b^2 +2c  α^2 +β^2 +γ^2 +δ^2 =2b^2   (α+β)^2 −2αβ+(γ+δ)^2 −2γδ=2b^2   (α+β)^2 −2(c)+(γ+δ)^2 −2(−c)=2b^2   (α+β)^2 +(γ+δ)^2 =2b^2   2b^2 =p^2 +q^2
$$\bullet{x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\: \\ $$$$\alpha+\beta=−{b}\:,\:\:\:\alpha\beta={c} \\ $$$$\bullet{x}^{\mathrm{2}} +{bx}−{c}=\mathrm{0} \\ $$$$\gamma+\delta=−{b}\:\:,\:\:\gamma\delta=−{c} \\ $$$$\alpha+\beta=\gamma+\delta\:\:,\:\alpha\beta=−\gamma\delta \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\mathrm{2}\alpha\beta \\ $$$$\:\:\:\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} ={b}^{\mathrm{2}} −\mathrm{2}{c} \\ $$$$\left(\gamma+\delta\right)^{\mathrm{2}} =\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} +\mathrm{2}\gamma\delta \\ $$$$\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} ={b}^{\mathrm{2}} +\mathrm{2}{c} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\delta^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta+\left(\gamma+\delta\right)^{\mathrm{2}} −\mathrm{2}\gamma\delta=\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\cancel{\mathrm{2}\left({c}\right)}+\left(\gamma+\delta\right)^{\mathrm{2}} −\cancel{\mathrm{2}\left(−{c}\right)}=\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} +\left(\gamma+\delta\right)^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$
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
This is just 2b^2 =b^2 +b^2  :)
$$\left.\mathcal{T}{his}\:{is}\:{just}\:\mathrm{2}{b}^{\mathrm{2}} ={b}^{\mathrm{2}} +{b}^{\mathrm{2}} \::\right) \\ $$

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