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Question Number 104191 by mathocean1 last updated on 19/Jul/20

Given (E):x^4 −10x^2 +q=0 U=x^2 so (E_((U)) )=U^2 −40U+q=0 . We suppose that E_((U)) has two roots such as r_1 <r_(2 ) . We give also that r_1 +r_2 =40 and r_1 ×r_2 =q. 1)The equation E has four positive solution. Determinate them such that these solutions form an arithmetical progression.

$${Given}\:\left({E}\right):{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +{q}=\mathrm{0} \\ $$ $${U}={x}^{\mathrm{2}} \:{so}\:\left({E}_{\left({U}\right)} \right)={U}^{\mathrm{2}} −\mathrm{40}{U}+{q}=\mathrm{0}\:. \\ $$ $${We}\:{suppose}\:{that}\:{E}_{\left({U}\right)} {has}\:{two}\:{roots} \\ $$ $${such}\:{as}\:{r}_{\mathrm{1}} <{r}_{\mathrm{2}\:} . \\ $$ $${We}\:{give}\:{also}\:{that}\:{r}_{\mathrm{1}} +{r}_{\mathrm{2}} =\mathrm{40}\:{and} \\ $$ $${r}_{\mathrm{1}} ×{r}_{\mathrm{2}} ={q}. \\ $$ $$\left.\mathrm{1}\right){The}\:{equation}\:{E}\:{has}\:{four}\:{positive} \\ $$ $${solution}.\:{Determinate}\:{them}\:{such} \\ $$ $${that}\:{these}\:{solutions}\:{form}\:{an}\: \\ $$ $${arithmetical}\:{progression}. \\ $$ $$ \\ $$

Answered by 1549442205PVT last updated on 20/Jul/20

Four roots of eqs.(E) are :±(√(r_1 )) ,±(√(r_2 )) Since they form an arithmetical progression and r_1 <r_2 ,they arrange from least to bigest values as:−(√r_2 ) ,−(√r_1 ) ,(√r_1 ),(√r_2 ) By the property of the arithmetical progression we get (√r_2 ) −(√r_1 )=2(√r_1 ) ⇔r_1 +r_2 −2(√(r_1 r_2 ))=4r_1 From the hypothesis r_1 +r_2 =40 we get 40−2(√(r_1 r_2 )) =4r_1 ⇔20−2r_1 =(√(r_1 (40−r_1 ))) ⇔400−80r_1 +4r_1 ^2 =r_1 (40−r_1 ) ⇔5r_1 ^2 −120r_1 +400=0⇔r_1 ^2 −24r_1 +80=0 Δ′=144−80=64⇒(√(Δ ))=8 ⇒r_1 =12±8⇒r_1 =4(r_1 =20 is rejected) ⇒r_2 =36.⇒the roots of eqs.(E) are x∈{−6;−2;2;6}

$$\mathrm{Four}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{eqs}.\left(\mathrm{E}\right)\:\mathrm{are}\::\pm\sqrt{\mathrm{r}_{\mathrm{1}} \:}\:,\pm\sqrt{\mathrm{r}_{\mathrm{2}} \:} \\ $$ $$\mathrm{Since}\:\mathrm{they}\:\mathrm{form}\:\mathrm{an}\:\mathrm{arithmetical} \\ $$ $$\mathrm{progression}\:\mathrm{and}\:\mathrm{r}_{\mathrm{1}} <\mathrm{r}_{\mathrm{2}} ,\mathrm{they}\:\mathrm{arrange}\:\mathrm{from}\:\mathrm{least} \\ $$ $$\mathrm{to}\:\mathrm{bigest}\:\mathrm{values}\:\mathrm{as}:−\sqrt{\mathrm{r}_{\mathrm{2}} }\:,−\sqrt{\mathrm{r}_{\mathrm{1}} }\:,\sqrt{\mathrm{r}_{\mathrm{1}} },\sqrt{\mathrm{r}_{\mathrm{2}} } \\ $$ $$\mathrm{By}\:\mathrm{the}\:\mathrm{property}\:\mathrm{of}\:\mathrm{the}\:\mathrm{arithmetical} \\ $$ $$\mathrm{progression}\:\mathrm{we}\:\mathrm{get} \\ $$ $$\sqrt{\mathrm{r}_{\mathrm{2}} }\:−\sqrt{\mathrm{r}_{\mathrm{1}} }=\mathrm{2}\sqrt{\mathrm{r}_{\mathrm{1}} }\:\Leftrightarrow\mathrm{r}_{\mathrm{1}} +\mathrm{r}_{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{r}_{\mathrm{1}} \mathrm{r}_{\mathrm{2}} }=\mathrm{4r}_{\mathrm{1}} \\ $$ $$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{r}_{\mathrm{1}} +\mathrm{r}_{\mathrm{2}} =\mathrm{40}\:\mathrm{we}\:\mathrm{get} \\ $$ $$\mathrm{40}−\mathrm{2}\sqrt{\mathrm{r}_{\mathrm{1}} \mathrm{r}_{\mathrm{2}} }\:=\mathrm{4r}_{\mathrm{1}} \Leftrightarrow\mathrm{20}−\mathrm{2r}_{\mathrm{1}} =\sqrt{\mathrm{r}_{\mathrm{1}} \left(\mathrm{40}−\mathrm{r}_{\mathrm{1}} \right)} \\ $$ $$\Leftrightarrow\mathrm{400}−\mathrm{80r}_{\mathrm{1}} +\mathrm{4r}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{r}_{\mathrm{1}} \left(\mathrm{40}−\mathrm{r}_{\mathrm{1}} \right) \\ $$ $$\Leftrightarrow\mathrm{5r}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{120r}_{\mathrm{1}} +\mathrm{400}=\mathrm{0}\Leftrightarrow\mathrm{r}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{24r}_{\mathrm{1}} +\mathrm{80}=\mathrm{0} \\ $$ $$\Delta'=\mathrm{144}−\mathrm{80}=\mathrm{64}\Rightarrow\sqrt{\Delta\:}=\mathrm{8} \\ $$ $$\Rightarrow\mathrm{r}_{\mathrm{1}} =\mathrm{12}\pm\mathrm{8}\Rightarrow\mathrm{r}_{\mathrm{1}} =\mathrm{4}\left(\mathrm{r}_{\mathrm{1}} =\mathrm{20}\:\mathrm{is}\:\mathrm{rejected}\right) \\ $$ $$\Rightarrow\mathrm{r}_{\mathrm{2}} =\mathrm{36}.\Rightarrow\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\:\mathrm{eqs}.\left(\mathrm{E}\right)\:\mathrm{are} \\ $$ $$\boldsymbol{\mathrm{x}}\in\left\{−\mathrm{6};−\mathrm{2};\mathrm{2};\mathrm{6}\right\} \\ $$

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