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Question Number 140973 by EnterUsername last updated on 14/May/21

If α and β are roots of the equation 2x^2 +ax+b=0, then one of the roots of the equation 2(αx+β)^2 + a(αx+β)+b=0 is (A) 0 (B) ((α+2b)/α^2 ) (C) ((aα+b)/(2α^2 )) (D) ((aα−2b)/(2α^2 ))

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2x}^{\mathrm{2}} +{a}\mathrm{x}+{b}=\mathrm{0}, \\ $$$$\mathrm{then}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}\left(\alpha\mathrm{x}+\beta\right)^{\mathrm{2}} + \\ $$$${a}\left(\alpha\mathrm{x}+\beta\right)+{b}=\mathrm{0}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\alpha+\mathrm{2}{b}}{\alpha^{\mathrm{2}} } \\ $$$$\left(\mathrm{C}\right)\:\frac{{a}\alpha+{b}}{\mathrm{2}\alpha^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{{a}\alpha−\mathrm{2}{b}}{\mathrm{2}\alpha^{\mathrm{2}} } \\ $$

Answered by TheSupreme last updated on 15/May/21

α=((−a+(√Δ))/4) β=((−a−(√Δ))/4) 2(αx+β)^2 +a(αx+β)+b=0 αx+β=α x=((α−β)/α) αx+β=β x=0 (A)

$$\alpha=\frac{−{a}+\sqrt{\Delta}}{\mathrm{4}} \\ $$$$\beta=\frac{−{a}−\sqrt{\Delta}}{\mathrm{4}} \\ $$$$\mathrm{2}\left(\alpha{x}+\beta\right)^{\mathrm{2}} +{a}\left(\alpha{x}+\beta\right)+{b}=\mathrm{0} \\ $$$$\alpha{x}+\beta=\alpha \\ $$$${x}=\frac{\alpha−\beta}{\alpha} \\ $$$$\alpha{x}+\beta=\beta \\ $$$${x}=\mathrm{0}\:\left({A}\right) \\ $$

Commented by EnterUsername last updated on 19/May/21

Thanks so much

$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much} \\ $$

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