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Question Number 43630 by peter frank last updated on 12/Sep/18

$$\mathrm{If}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$x^3-12x^2+39x-28=0\mathrm{are}\mathrm{in}\mathrm{AP},\mathrm{then}$$$$\mathrm{their}\mathrm{common}\mathrm{difference}\mathrm{will}\mathrm{be}$$

Answered by $@ty@m last updated on 13/Sep/18

$$(\alpha -\beta )+\alpha +(\alpha +\beta )=12$$$$\Rightarrow 3\alpha =12$$$$\Rightarrow \alpha =4...\left(1\right)$$$$(\alpha -\beta ).\alpha .(\alpha +\beta )=28$$$$\Rightarrow \alpha .({\alpha }^2-{\beta }^2)=28$$$$\Rightarrow 4\times (4^2-{\beta }^2)=28$$$$\Rightarrow 16-{\beta }^2=7$$$$\Rightarrow {\beta }^2=9$$$$\beta =\pm 3Ans.$$

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