Question Number 93705 by i jagooll last updated on 14/May/20
$$\mathrm{find}\:\mathrm{real}\:\mathrm{q}\:\mathrm{so}\:\mathrm{that}\:\mathrm{x}^{\mathrm{4}} −\mathrm{40x}^{\mathrm{2}} +\mathrm{q}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{four}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{forming} \\ $$$$\mathrm{AP}.\: \\ $$
Commented by PRITHWISH SEN 2 last updated on 14/May/20
$$\mathrm{let}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\boldsymbol{\mathrm{a}}−\mathrm{3}\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{a}}+\mathrm{3}\boldsymbol{\mathrm{b}} \\ $$$$\therefore\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots} \\ $$$$\mathrm{4}\boldsymbol{\mathrm{a}}=\mathrm{0}\Rightarrow\boldsymbol{\mathrm{a}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{product}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{roots}}\:\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{time}} \\ $$$$−\mathrm{10}\boldsymbol{\mathrm{b}}^{\mathrm{2}} =\mathrm{40}\:\Rightarrow\mathrm{b}=\pm\mathrm{2} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\:−\mathrm{6},−\mathrm{2},\mathrm{2},\mathrm{6} \\ $$$$\therefore\boldsymbol{\mathrm{q}}=\mathrm{36}×\mathrm{4}=\mathrm{144}.\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
Commented by i jagooll last updated on 14/May/20
correct sir. but my way is different
Commented by PRITHWISH SEN 2 last updated on 14/May/20
$$\mathrm{then}\:\mathrm{post}\:\mathrm{it}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{also}\:\mathrm{lots}\:\mathrm{to}\:\mathrm{learn}. \\ $$