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.

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}