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Question Number 51691 by peter frank last updated on 29/Dec/18

Answered by afachri last updated on 29/Dec/18

$$\mathrm{suppose}\gamma =U_1;\beta =U_2;\alpha =U_3$$$$\mathit{\gamma },\mathit{\beta }=\frac{-a\pm \sqrt{a^2-4b}}{2}$$$$\mathit{\beta }-\mathit{\gamma }=\frac{(-a-\sqrt{a^2-4b})-(-a+\sqrt{a^2-4b})}{2}$$$$\mathit{\beta }-\mathit{\gamma }=-\sqrt{a^2-4b}\mathbf{or}\beta -\gamma =+\sqrt{a^2-4b}$$$$\mathbf{in}\mathbf{A}.\mathbf{P}:$$$$\beta -\gamma =\alpha -\beta$$$$\pm \sqrt{a^2-4b}=\alpha -\frac{-a\pm \sqrt{a^2-4b}}{2}$$$$\alpha =\pm \sqrt{a^2-4b}+\frac{-a\pm \sqrt{a^2-4b}}{2}$$$$\alpha =\frac{\pm 2\sqrt{a^2-4b}-a\pm \sqrt{a^2-4b}}{2}$$$$\alpha =\frac{-a\pm 3\sqrt{a^2-4b}}{2}$$$$\mathbf{the}\mathbf{roots}\mathbf{of}{y}^2+ay+(9b-2a^2)=0$$$$y_{1,2}=\frac{-a\pm \sqrt{a^2-4(9b-2a^2)}}{2}$$$$=\frac{-a\pm \sqrt{a^2+8a^2-36b}}{2}$$$$=\frac{-a\pm \sqrt{9a^2-36b}}{2}$$$$=\frac{-a\pm \sqrt{9(a^2-4b)}}{2}$$$$=\frac{-a\pm 3\sqrt{(a^2-4b)}}{2}$$$$\mathbf{we}\mathbf{get}\mathbf{same}\mathbf{product}\mathbf{for}\mathit{\alpha }\mathbf{and}{\mathit{y}}_{1,2}$$$$$$

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