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.


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

$G i v e n (E) : x^{4} - 10 x^{2} + q = 0$ $U = x^{2} s o (E_{(U)}) = U^{2} - 40 U + q = 0 .$ $W e s u p p o s e t h a t E_{(U)} h a s t w o r o o t s$ $s u c h a s r_{1} < r_{2} .$ $W e g i v e a l s o t h a t r_{1} + r_{2} = 40 a n d$ $r_{1} \times r_{2} = q .$ $1) T h e e q u a t i o n E h a s f o u r p o s i t i v e$ $s o l u t i o n . D e t e r m i n a t e t h e m s u c h$ $t h a t t h e s e s o l u t i o n s f o r m a n$ $a r i t h m e t i c a l p r o g r e s s i o n .$
	Answered by 1549442205PVT last updated on 20/Jul/20
	$Four roots of eqs.(E) are :±(√(r_1 )) ,±(√(r_2 )) Since they form an arithmetical progression and r_1 <r_2 ,they arrange from least to bigest values as:−(√r_2 ) ,−(√r_1 ) ,(√r_1 ),(√r_2 ) By the property of the arithmetical progression we get (√r_2 ) −(√r_1 )=2(√r_1 ) ⇔r_1 +r_2 −2(√(r_1 r_2 ))=4r_1 From the hypothesis r_1 +r_2 =40 we get 40−2(√(r_1 r_2 )) =4r_1 ⇔20−2r_1 =(√(r_1 (40−r_1 ))) ⇔400−80r_1 +4r_1 ^2 =r_1 (40−r_1 ) ⇔5r_1 ^2 −120r_1 +400=0⇔r_1 ^2 −24r_1 +80=0 Δ′=144−80=64⇒(√(Δ ))=8 ⇒r_1 =12±8⇒r_1 =4(r_1 =20 is rejected) ⇒r_2 =36.⇒the roots of eqs.(E) are x∈{−6;−2;2;6}$
	$Four roots of eqs . (E) are : \pm \sqrt{r_{1}}, \pm \sqrt{r_{2}}$ $Since they form an arithmetical$ $progression and r_{1} < r_{2}, they arrange from least$ $to bigest values as : - \sqrt{r_{2}}, - \sqrt{r_{1}}, \sqrt{r_{1}}, \sqrt{r_{2}}$ $By the property of the arithmetical$ $progression we get$ $\sqrt{r_{2}} - \sqrt{r_{1}} = 2 \sqrt{r_{1}} \Leftrightarrow r_{1} + r_{2} - 2 \sqrt{r_{1} r_{2}} = {4 r}_{1}$ $From the hypothesis r_{1} + r_{2} = 40 we get$ $40 - 2 \sqrt{r_{1} r_{2}} = {4 r}_{1} \Leftrightarrow 20 - {2 r}_{1} = \sqrt{r_{1} (40 - r_{1})}$ $\Leftrightarrow 400 - {80 r}_{1} + {4 r}_{1}^{2} = r_{1} (40 - r_{1})$ $\Leftrightarrow {5 r}_{1}^{2} - {120 r}_{1} + 400 = 0 \Leftrightarrow r_{1}^{2} - {24 r}_{1} + 80 = 0$ $Δ^{'} = 144 - 80 = 64 \Rightarrow \sqrt{Δ} = 8$ $\Rightarrow r_{1} = 12 \pm 8 \Rightarrow r_{1} = 4 (r_{1} = 20 is rejected)$ $\Rightarrow r_{2} = 36 . \Rightarrow the roots of eqs . (E) are$ $x \in {- 6; - 2; 2; 6}$
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