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If-the-roots-of-the-equation-x-3-12x-2-39x-28-0-are-in-AP-then-their-common-difference-will-be-




Question Number 43630 by peter frank last updated on 12/Sep/18
If  the roots of the equation   x^3 −12x^2 +39x−28=0  are in AP, then  their common difference will be
$$\mathrm{If}\:\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$${x}^{\mathrm{3}} −\mathrm{12}{x}^{\mathrm{2}} +\mathrm{39}{x}−\mathrm{28}=\mathrm{0}\:\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP},\:\mathrm{then} \\ $$$$\mathrm{their}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{will}\:\mathrm{be} \\ $$
Answered by $@ty@m last updated on 13/Sep/18
(α−β)+α+(α+β)=12  ⇒3α=12  ⇒α=4 ...(1)  (α−β).α.(α+β)=28  ⇒α.(α^2 −β^2 )=28  ⇒4×(4^2 −β^2 )=28  ⇒16−β^2 =7  ⇒β^2 =9  β=±3 Ans.
$$\left(\alpha−\beta\right)+\alpha+\left(\alpha+\beta\right)=\mathrm{12} \\ $$$$\Rightarrow\mathrm{3}\alpha=\mathrm{12} \\ $$$$\Rightarrow\alpha=\mathrm{4}\:…\left(\mathrm{1}\right) \\ $$$$\left(\alpha−\beta\right).\alpha.\left(\alpha+\beta\right)=\mathrm{28} \\ $$$$\Rightarrow\alpha.\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)=\mathrm{28} \\ $$$$\Rightarrow\mathrm{4}×\left(\mathrm{4}^{\mathrm{2}} −\beta^{\mathrm{2}} \right)=\mathrm{28} \\ $$$$\Rightarrow\mathrm{16}−\beta^{\mathrm{2}} =\mathrm{7} \\ $$$$\Rightarrow\beta^{\mathrm{2}} =\mathrm{9} \\ $$$$\beta=\pm\mathrm{3}\:{Ans}. \\ $$

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