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bemath-What-is-m-so-the-roots-of-x-4-m-2-x-2-9-0-are-in-AP-

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Question Number 106880 by bemath last updated on 07/Aug/20
^(@bemath@) What is m so the roots of x^4 −(m+2)x^2 +9=0 are in AP
@bemath@Whatismsotherootsofx4−(m+2)x2+9=0areinAP
Commented by Her_Majesty last updated on 07/Aug/20
so the roots of ... what? this doesn′t make sense, it′s not a full sentence
sotherootsof…what?thisdoesn′tmakesense,it′snotafullsentence
Commented by bemath last updated on 07/Aug/20
yes..you are right
yes..youareright
Answered by john santu last updated on 07/Aug/20
^(◊JS⧫) say the roots are η , η+d, η+2d, η+3d By Vieta′s rule { ((4η+6d = 0 → d=−(2/3)η)),((η(η+d)(η+2d)(η+3d)=9)) :} ⇒η(η−(2/3)η)(η−(4/3)η)(η−(6/3)η)=9 η((1/3)η)(−(1/3)η)(−η)=9 η^4 =81 → { ((η=3)),((η=−3)) :} case(1) for η=3 substute to quartic equation ⇒3^4 −(m+2)3^2 +9=0 9−(m+2)+1=0 ⇒m=8 case(2) for η=−3 ⇒equal to η=3 therefore the value of m is 8.
◊JS⧫saytherootsareη,η+d,η+2d,η+3dByVieta′srule{4η+6d=0→d=−23ηη(η+d)(η+2d)(η+3d)=9⇒η(η−23η)(η−43η)(η−63η)=9η(13η)(−13η)(−η)=9η4=81→{η=3η=−3case(1)forη=3substutetoquarticequation⇒34−(m+2)32+9=09−(m+2)+1=0⇒m=8case(2)forη=−3⇒equaltoη=3thereforethevalueofmis8.
Commented by Her_Majesty last updated on 07/Aug/20
how do you know what is asked? coffee cup reading? telepathy?
howdoyouknowwhatisasked?coffeecupreading?telepathy?
Commented by john santu last updated on 07/Aug/20
by feeling
byfeeling
Answered by 1549442205PVT last updated on 07/Aug/20
The equation x^4 −(m+2)x^2 +9=0 has roots establish an AP if and only if two following conditions is satisfied:(set y=t^2 ) i)y^2 −(m+2)y^2 −(m+2)y+9=0(∗) has two positive roots y_1 ,y_2 .Since y_1 y_2 =9>0 ,this is equavalent to { ((Δ=(m+2)^2 −36>0)),((y_1 +y_2 =m+2>0)) :}⇔ { ((m^2 +4m−32>0)),((m+2>0)) :} ⇔ { ((((m−4)(m+8)>0)),((m+2>0)) :}⇔ { ((m∈(−∞;−8)∪(4;+∞))),((m>−2)) :} ⇔m∈(4;+∞)(∗∗) ii)Suppose that y_1 <y_2 and x_1 ,x_2 ,x_3 ,x_4 establish an arithmetic progression consecutively.Then we have: x_1 =−(√y_2 ) ,x_2 =−(√y_1 ) ,x_3 =(√y_1 ) ,x_4 =(√y_2 ) x_2 −x_1 =x_3 −x_2 =x_4 −x_3 =d ⇒ { ((2x_2 =x_1 +x_3 )),((2x_3 =x_2 +x_4 )) :}⇔ { ((4x_2 =2x_1 +2x_3 )),((2x_3 =x_2 +x_4 )) :} ⇒Adding up we get 4x_2 =2x_1 +x_2 +x_4 ⇔3x_2 =x_(1 ) (since x_1 +x_4 =x_2 +x_3 =0)(1) .On ther other hands, by Vieta′s theorem we have: y_1 +y_2 =m+2=x_2 ^2 +x_1 ^2 =x_3 ^2 +x_4 ^4 (2) y_1 y_2 =9=(x_1 x_2 )^2 =(x_3 x_4 )^2 ⇔x_1 x_2 =x_3 x_4 =3.From (1)we get 3x_2 ^2 =3⇒x_2 ^2 =1⇒x_2 =−1⇒x_1 =−3 Replace into (2) we obtain:m+2=10 ⇒m=8.This value satisfy the condition(∗∗) Consequently,The equation x^4 −(m+2)x^2 +9=0 has roots establish an arithmetic progression if an only if m=8
Theequationx4−(m+2)x2+9=0hasrootsestablishanAPifandonlyiftwofollowingconditionsissatisfied:(sety=t2)i)y2−(m+2)y2−(m+2)y+9=0(∗)hastwopositiverootsy1,y2.Sincey1y2=9>0,thisisequavalentto{Δ=(m+2)2−36>0y1+y2=m+2>0⇔{m2+4m−32>0m+2>0⇔{((m−4)(m+8)>0m+2>0⇔{m∈(−∞;−8)∪(4;+∞)m>−2⇔m∈(4;+∞)(∗∗)ii)Supposethaty1<y2andx1,x2,x3,x4establishanarithmeticprogressionconsecutively.Thenwehave:x1=−y2,x2=−y1,x3=y1,x4=y2x2−x1=x3−x2=x4−x3=d⇒{2x2=x1+x32x3=x2+x4⇔{4x2=2x1+2x32x3=x2+x4⇒Addingupweget4x2=2x1+x2+x4⇔3x2=x1(sincex1+x4=x2+x3=0)(1).Ontherotherhands,byVieta′stheoremwehave:y1+y2=m+2=x22+x12=x32+x44(2)y1y2=9=(x1x2)2=(x3x4)2⇔x1x2=x3x4=3.From(1)weget3x22=3⇒x22=1⇒x2=−1⇒x1=−3Replaceinto(2)weobtain:m+2=10⇒m=8.Thisvaluesatisfythecondition(∗∗)Consequently,Theequationx4−(m+2)x2+9=0hasrootsestablishanarithmeticprogressionifanonlyifm=8
Answered by $@y@m last updated on 08/Aug/20
Let the roots be a−3d, a−d,a+d,a+3d Then, 4a=0 & (−3d)(−d)(d)(3d)=9 ⇒ a=0 & d=1 ∴ the roors are −3, −1, 1 & 3 Now, −(m+2)={(−3)+(−1)}.(1+3)+3+3 ⇒m+2=−16+6 m=8 OR, The equation whose roots are −3, −1, 1 & 3 is (x^2 −9)(x^2 −1)=0 ⇒x^4 −10x^2 +9=0 Equating the coefficient ofx^2 , we get m+2=10⇒m=8
Lettherootsbea−3d,a−d,a+d,a+3dThen,4a=0&(−3d)(−d)(d)(3d)=9⇒a=0&d=1∴theroorsare−3,−1,1&3Now,−(m+2)={(−3)+(−1)}.(1+3)+3+3⇒m+2=−16+6m=8OR,Theequationwhoserootsare−3,−1,1&3is(x2−9)(x2−1)=0⇒x4−10x2+9=0Equatingthecoefficientofx2,wegetm+2=10⇒m=8
Commented by Rasheed.Sindhi last updated on 08/Aug/20
V Nice!
VNice!VNice!
Commented by $@y@m last updated on 08/Aug/20
Thanks!
Thanks!Thanks!

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