Question Number 98058 by PengagumRahasiamu last updated on 11/Jun/20
$$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} −{cx}+{d}=\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{fourth}\:\mathrm{powers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots} \\ $$$$\mathrm{of}\:{x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0},\:\mathrm{where}\:{a}^{\mathrm{2}} >\mathrm{4}{b}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} −\mathrm{4}{bx}+\mathrm{2}{b}^{\mathrm{2}} −{c}=\mathrm{0}\:\mathrm{will}\:\mathrm{be} \\ $$
Answered by Rio Michael last updated on 11/Jun/20
![Both real(one is positive and the other negative) see how: let x^2 + ax + b = 0 have roots α and β then the roots of x^2 −cx +d = 0 are α^4 and β^4 α + β = −a and αβ = b also α^4 + β^4 = c and α^4 β^4 = d ⇒ b^4 = d and α^4 + β^4 = c (α^2 +β^2 )^2 −2(αβ)^2 = c [(α + β)^2 −2αβ]−2(αβ)^2 = c (a^2 −2b^2 )−2b^2 = c ⇒ 2b^2 + c = (a^2 −2b)^2 2b^2 −c = 4a^2 b−a^2 = a^2 (4b−a^2 ) now for x^2 −4bx + 2b^2 −c = 0 discriminantD = (4b)^2 −4(1)(2b^2 −c) = 16b^2 −8b^2 + 4c = 8b^2 + 4c = 4(2b^2 + c) = 4(a^2 −2b)^2 >0 ⇒ real roots f(0) = 2b^2 −c = a^2 (4b−a^2 ) < 0 ⇒ roots have opposite sign](https://www.tinkutara.com/question/Q98066.png)
$$\mathrm{Both}\:\mathrm{real}\left(\mathrm{one}\:\mathrm{is}\:\mathrm{positive}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{negative}\right) \\ $$$$\mathrm{see}\:\mathrm{how}: \\ $$$$\mathrm{let}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\mathrm{0}\:\mathrm{have}\:\mathrm{roots}\:\alpha\:\mathrm{and}\:\beta \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} −{cx}\:+{d}\:=\:\mathrm{0}\:\mathrm{are}\:\:\alpha^{\mathrm{4}} \:\mathrm{and}\:\beta^{\mathrm{4}} \\ $$$$\alpha\:+\:\beta\:=\:−{a}\:\mathrm{and}\:\alpha\beta\:=\:{b} \\ $$$$\mathrm{also}\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:=\:{c}\:\mathrm{and}\:\alpha^{\mathrm{4}} \beta^{\mathrm{4}} \:=\:{d} \\ $$$$\Rightarrow\:{b}^{\mathrm{4}} \:=\:{d}\:\mathrm{and}\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:=\:{c} \\ $$$$\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta\right)^{\mathrm{2}} \:=\:{c} \\ $$$$\:\left[\left(\alpha\:+\:\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\right]−\mathrm{2}\left(\alpha\beta\right)^{\mathrm{2}} \:=\:{c} \\ $$$$\:\:\left({a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \right)−\mathrm{2}{b}^{\mathrm{2}} \:=\:{c}\:\:\Rightarrow\:\mathrm{2}{b}^{\mathrm{2}} \:+\:{c}\:=\:\left({a}^{\mathrm{2}} −\mathrm{2}{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −{c}\:=\:\mathrm{4}{a}^{\mathrm{2}} {b}−{a}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{a}^{\mathrm{2}} \left(\mathrm{4}{b}−{a}^{\mathrm{2}} \right) \\ $$$$\mathrm{now}\:\mathrm{for}\:{x}^{\mathrm{2}} −\mathrm{4}{bx}\:+\:\mathrm{2}{b}^{\mathrm{2}} −{c}\:=\:\mathrm{0}\: \\ $$$$\mathrm{discriminant}{D}\:=\:\left(\mathrm{4}{b}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{2}{b}^{\mathrm{2}} −{c}\right) \\ $$$$\:\:=\:\mathrm{16}{b}^{\mathrm{2}} −\mathrm{8}{b}^{\mathrm{2}} \:+\:\mathrm{4}{c}\: \\ $$$$\:\:=\:\mathrm{8}{b}^{\mathrm{2}} \:+\:\mathrm{4}{c}\: \\ $$$$\:\:=\:\mathrm{4}\left(\mathrm{2}{b}^{\mathrm{2}} \:+\:{c}\right) \\ $$$$\:\:\:=\:\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{2}{b}\right)^{\mathrm{2}} \:>\mathrm{0}\:\Rightarrow\:\mathrm{real}\:\mathrm{roots} \\ $$$${f}\left(\mathrm{0}\right)\:=\:\mathrm{2}{b}^{\mathrm{2}} −{c}\:=\:{a}^{\mathrm{2}} \left(\mathrm{4}{b}−{a}^{\mathrm{2}} \right)\:<\:\mathrm{0}\:\Rightarrow\:\mathrm{roots}\:\mathrm{have}\:\mathrm{opposite}\:\mathrm{sign}\: \\ $$