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

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

Answered by TheSupreme last updated on 15/May/21

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

Commented by EnterUsername last updated on 19/May/21

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

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