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find-the-relation-between-a-b-c-and-d-if-the-roots-of-the-eqution-ax-3-bx-2-cx-d-0-are-in-i-A-P-ii-G-P-

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Question Number 9873 by j.masanja06@gmail.com last updated on 11/Jan/17
find the relation between a,b,c and d if the roots of the eqution ax^3 + bx^2 +cx+d=0 are in; (i)A.P (ii)G.P
$$\mathrm{find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{and}\:\:\mathrm{d} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{eqution}\: \\ $$$$\mathrm{ax}^{\mathrm{3}} \:+\:\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}+\mathrm{d}=\mathrm{0}\:\:\mathrm{are}\:\mathrm{in}; \\ $$$$\left(\mathrm{i}\right)\mathrm{A}.\mathrm{P} \\ $$$$\left(\mathrm{ii}\right)\mathrm{G}.\mathrm{P} \\ $$
Answered by bansal22luvi@gmail.com last updated on 11/Jan/17
when the roots are in AP let the roots be (α−δ) . α . (α+δ) 3α = −(b/a) α=((−b)/(3a))..............(1) (α−δ)α(α+δ)=−(d/a) α^3 −αδ^2 = −(d/a) −(b^3 /(27a^3 ))+(b/(3a))δ^2 =−(d/a)(by eqn 1) ((−b^3 +9a^2 bδ^2 )/(27a^3 ))=−(d/a) δ^2 =((−3a^2 d+b^3 )/(a^2 b))........(2) α(α−δ)+(α−δ)(α+δ)+α(α+δ)=(c/a) 3α^2 −δ^2 =(c/a) (b^2 /(3a^2 ))−((b^3 −3a^2 d)/(a^2 b))=(c/a)(by eqn 1 and 2) ((−2b^3 +9a^2 d)/(3a^2 b))=(c/a) 9a^2 d=3abc+2b^3 same as done when the roots are in GP Kush
$${when}\:{the}\:{roots}\:{are}\:{in}\:{AP} \\ $$$${let}\:{the}\:{roots}\:{be}\:\left(\alpha−\delta\right)\:.\:\alpha\:.\:\left(\alpha+\delta\right) \\ $$$$\mathrm{3}\alpha\:=\:−\frac{{b}}{{a}} \\ $$$$\alpha=\frac{−{b}}{\mathrm{3}{a}}…………..\left(\mathrm{1}\right) \\ $$$$\left(\alpha−\delta\right)\alpha\left(\alpha+\delta\right)=−\frac{{d}}{{a}} \\ $$$$\alpha^{\mathrm{3}} −\alpha\delta^{\mathrm{2}} =\:−\frac{{d}}{{a}} \\ $$$$−\frac{{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }+\frac{{b}}{\mathrm{3}{a}}\delta^{\mathrm{2}} =−\frac{{d}}{{a}}\left({by}\:{eqn}\:\mathrm{1}\right) \\ $$$$\frac{−{b}^{\mathrm{3}} +\mathrm{9}{a}^{\mathrm{2}} {b}\delta^{\mathrm{2}} }{\mathrm{27}{a}^{\mathrm{3}} }=−\frac{{d}}{{a}} \\ $$$$\delta^{\mathrm{2}} =\frac{−\mathrm{3}{a}^{\mathrm{2}} {d}+{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} {b}}……..\left(\mathrm{2}\right) \\ $$$$\alpha\left(\alpha−\delta\right)+\left(\alpha−\delta\right)\left(\alpha+\delta\right)+\alpha\left(\alpha+\delta\right)=\frac{{c}}{{a}} \\ $$$$\mathrm{3}\alpha^{\mathrm{2}} −\delta^{\mathrm{2}} =\frac{{c}}{{a}} \\ $$$$\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} }−\frac{{b}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} {d}}{{a}^{\mathrm{2}} {b}}=\frac{{c}}{{a}}\left({by}\:{eqn}\:\mathrm{1}\:{and}\:\mathrm{2}\right) \\ $$$$\frac{−\mathrm{2}{b}^{\mathrm{3}} +\mathrm{9}{a}^{\mathrm{2}} {d}}{\mathrm{3}{a}^{\mathrm{2}} {b}}=\frac{{c}}{{a}} \\ $$$$\mathrm{9}{a}^{\mathrm{2}} {d}=\mathrm{3}{abc}+\mathrm{2}{b}^{\mathrm{3}} \\ $$$${same}\:{as}\:{done}\:{when}\:{the}\:{roots}\:{are}\:{in}\:{GP} \\ $$$$\boldsymbol{{Kush}} \\ $$

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