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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
$$\:\:\:\:\:\:\:\:\:\:\overset{@\mathrm{bemath}@} {\:} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{m}\:\mathrm{so}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{4}} \:−\left(\mathrm{m}+\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{AP} \\ $$
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
$${so}\:{the}\:{roots}\:{of}\:…\:{what}?\:{this}\:{doesn}'{t}\:{make} \\ $$$${sense},\:{it}'{s}\:{not}\:{a}\:{full}\:{sentence} \\ $$
Commented by bemath last updated on 07/Aug/20
yes..you are right
$$\mathrm{yes}..\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$
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.
$$\:\:\:\:\:\overset{\lozenge\mathrm{JS}\blacklozenge} {\:} \\ $$$$\mathrm{say}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\eta\:,\:\eta+\mathrm{d},\:\eta+\mathrm{2d}, \\ $$$$\eta+\mathrm{3d} \\ $$$$\mathrm{By}\:\mathrm{Vieta}'\mathrm{s}\:\mathrm{rule}\: \\ $$$$\begin{cases}{\mathrm{4}\eta+\mathrm{6d}\:=\:\mathrm{0}\:\rightarrow\:\mathrm{d}=−\frac{\mathrm{2}}{\mathrm{3}}\eta}\\{\eta\left(\eta+\mathrm{d}\right)\left(\eta+\mathrm{2d}\right)\left(\eta+\mathrm{3d}\right)=\mathrm{9}}\end{cases} \\ $$$$\Rightarrow\eta\left(\eta−\frac{\mathrm{2}}{\mathrm{3}}\eta\right)\left(\eta−\frac{\mathrm{4}}{\mathrm{3}}\eta\right)\left(\eta−\frac{\mathrm{6}}{\mathrm{3}}\eta\right)=\mathrm{9} \\ $$$$\eta\left(\frac{\mathrm{1}}{\mathrm{3}}\eta\right)\left(−\frac{\mathrm{1}}{\mathrm{3}}\eta\right)\left(−\eta\right)=\mathrm{9} \\ $$$$\eta^{\mathrm{4}} =\mathrm{81}\:\rightarrow\begin{cases}{\eta=\mathrm{3}}\\{\eta=−\mathrm{3}}\end{cases} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{for}\:\eta=\mathrm{3} \\ $$$$\mathrm{substute}\:\mathrm{to}\:\mathrm{quartic}\:\mathrm{equation} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{4}} −\left(\mathrm{m}+\mathrm{2}\right)\mathrm{3}^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{9}−\left(\mathrm{m}+\mathrm{2}\right)+\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{m}=\mathrm{8} \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\:\mathrm{for}\:\eta=−\mathrm{3} \\ $$$$\Rightarrow\mathrm{equal}\:\mathrm{to}\:\eta=\mathrm{3}\: \\ $$$$\mathrm{therefore}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{m}\:\mathrm{is}\:\mathrm{8}. \\ $$
Commented by Her_Majesty last updated on 07/Aug/20
how do you know what is asked? coffee cup reading? telepathy?
$${how}\:{do}\:{you}\:{know}\:{what}\:{is}\:{asked}? \\ $$$${coffee}\:{cup}\:{reading}?\:{telepathy}? \\ $$
Commented by john santu last updated on 07/Aug/20
by feeling
$$\mathrm{by}\:\mathrm{feeling} \\ $$
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
$$\mathrm{The}\:\mathrm{equation}\:\:\mathrm{x}^{\mathrm{4}} \:−\left(\mathrm{m}+\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{roots}\:\mathrm{establish}\:\mathrm{an}\:\mathrm{AP} \\ $$$$\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\:\mathrm{two}\:\mathrm{following}\:\mathrm{conditions} \\ $$$$\mathrm{is}\:\mathrm{satisfied}:\left(\mathrm{set}\:\mathrm{y}=\mathrm{t}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{i}\right)\mathrm{y}^{\mathrm{2}} −\left(\mathrm{m}+\mathrm{2}\right)\mathrm{y}^{\mathrm{2}} −\left(\mathrm{m}+\mathrm{2}\right)\mathrm{y}+\mathrm{9}=\mathrm{0}\left(\ast\right)\: \\ $$$$\:\mathrm{has}\:\mathrm{two}\:\mathrm{positive}\:\mathrm{roots}\:\mathrm{y}_{\mathrm{1}} ,\mathrm{y}_{\mathrm{2}} .\mathrm{Since} \\ $$$$\mathrm{y}_{\mathrm{1}} \mathrm{y}_{\mathrm{2}} =\mathrm{9}>\mathrm{0}\:,\mathrm{this}\:\mathrm{is}\:\mathrm{equavalent}\:\mathrm{to} \\ $$$$\:\:\begin{cases}{\Delta=\left(\mathrm{m}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{36}>\mathrm{0}}\\{\mathrm{y}_{\mathrm{1}} +\mathrm{y}_{\mathrm{2}} =\mathrm{m}+\mathrm{2}>\mathrm{0}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{m}^{\mathrm{2}} +\mathrm{4m}−\mathrm{32}>\mathrm{0}}\\{\mathrm{m}+\mathrm{2}>\mathrm{0}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\left(\left(\mathrm{m}−\mathrm{4}\right)\left(\mathrm{m}+\mathrm{8}\right)>\mathrm{0}\right.}\\{\mathrm{m}+\mathrm{2}>\mathrm{0}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{m}\in\left(−\infty;−\mathrm{8}\right)\cup\left(\mathrm{4};+\infty\right)}\\{\mathrm{m}>−\mathrm{2}}\end{cases} \\ $$$$\Leftrightarrow\boldsymbol{\mathrm{m}}\in\left(\mathrm{4};+\infty\right)\left(\ast\ast\right) \\ $$$$\left.\mathrm{ii}\right)\mathrm{Suppose}\:\mathrm{that}\:\mathrm{y}_{\mathrm{1}} <\mathrm{y}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{x}_{\mathrm{3}} ,\mathrm{x}_{\mathrm{4}} \\ $$$$\mathrm{establish}\:\mathrm{an}\:\mathrm{arithmetic}\:\mathrm{progression}\: \\ $$$$\mathrm{consecutively}.\mathrm{Then}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\mathrm{x}_{\mathrm{1}} =−\sqrt{\mathrm{y}_{\mathrm{2}} }\:,\mathrm{x}_{\mathrm{2}} =−\sqrt{\mathrm{y}_{\mathrm{1}} }\:,\mathrm{x}_{\mathrm{3}} =\sqrt{\mathrm{y}_{\mathrm{1}} }\:,\mathrm{x}_{\mathrm{4}} =\sqrt{\mathrm{y}_{\mathrm{2}} } \\ $$$$\mathrm{x}_{\mathrm{2}} −\mathrm{x}_{\mathrm{1}} =\mathrm{x}_{\mathrm{3}} −\mathrm{x}_{\mathrm{2}} =\mathrm{x}_{\mathrm{4}} −\mathrm{x}_{\mathrm{3}} =\mathrm{d} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2x}_{\mathrm{2}} =\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{3}} }\\{\mathrm{2x}_{\mathrm{3}} =\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{4}} }\end{cases}\Leftrightarrow\begin{cases}{\mathrm{4x}_{\mathrm{2}} =\mathrm{2x}_{\mathrm{1}} +\mathrm{2x}_{\mathrm{3}} }\\{\mathrm{2x}_{\mathrm{3}} =\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{4}} }\end{cases} \\ $$$$\Rightarrow\mathrm{Adding}\:\mathrm{up}\:\mathrm{we}\:\mathrm{get}\:\mathrm{4x}_{\mathrm{2}} =\mathrm{2x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{4}} \\ $$$$\Leftrightarrow\mathrm{3x}_{\mathrm{2}} =\mathrm{x}_{\mathrm{1}\:} \left(\mathrm{since}\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{4}} =\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} =\mathrm{0}\right)\left(\mathrm{1}\right) \\ $$$$.\mathrm{On}\:\mathrm{ther}\:\mathrm{other}\:\mathrm{hands}, \\ $$$$\mathrm{by}\:\mathrm{Vieta}'\mathrm{s}\:\mathrm{theorem}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\mathrm{y}_{\mathrm{1}} +\mathrm{y}_{\mathrm{2}} =\mathrm{m}+\mathrm{2}=\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{x}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{4}} ^{\mathrm{4}} \left(\mathrm{2}\right) \\ $$$$\mathrm{y}_{\mathrm{1}} \mathrm{y}_{\mathrm{2}} =\mathrm{9}=\left(\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \right)^{\mathrm{2}} =\left(\mathrm{x}_{\mathrm{3}} \mathrm{x}_{\mathrm{4}} \right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} =\mathrm{x}_{\mathrm{3}} \mathrm{x}_{\mathrm{4}} =\mathrm{3}.\mathrm{From}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{3x}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{3}\Rightarrow\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{1}\Rightarrow\mathrm{x}_{\mathrm{2}} =−\mathrm{1}\Rightarrow\mathrm{x}_{\mathrm{1}} =−\mathrm{3} \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{obtain}:\mathrm{m}+\mathrm{2}=\mathrm{10} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{8}.\mathrm{This}\:\mathrm{value}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{condition}\left(\ast\ast\right) \\ $$$$\mathrm{Consequently},\mathrm{The}\:\mathrm{equation}\:\:\mathrm{x}^{\mathrm{4}} \:−\left(\mathrm{m}+\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{roots}\:\mathrm{establish}\:\mathrm{an}\:\mathrm{arithmetic} \\ $$$$\mathrm{progression}\:\mathrm{if}\:\mathrm{an}\:\mathrm{only}\:\mathrm{if}\:\mathrm{m}=\mathrm{8} \\ $$
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
$${Let}\:{the}\:{roots}\:{be} \\ $$$${a}−\mathrm{3}{d},\:{a}−{d},{a}+{d},{a}+\mathrm{3}{d} \\ $$$${Then}, \\ $$$$\mathrm{4}{a}=\mathrm{0}\:\:\:\&\:\left(−\mathrm{3}{d}\right)\left(−{d}\right)\left({d}\right)\left(\mathrm{3}{d}\right)=\mathrm{9} \\ $$$$\Rightarrow\:{a}=\mathrm{0}\:\&\:{d}=\mathrm{1} \\ $$$$\therefore\:{the}\:{roors}\:{are} \\ $$$$−\mathrm{3},\:−\mathrm{1},\:\mathrm{1}\:\&\:\mathrm{3} \\ $$$${Now},\:−\left({m}+\mathrm{2}\right)=\left\{\left(−\mathrm{3}\right)+\left(−\mathrm{1}\right)\right\}.\left(\mathrm{1}+\mathrm{3}\right)+\mathrm{3}+\mathrm{3} \\ $$$$\:\Rightarrow{m}+\mathrm{2}=−\mathrm{16}+\mathrm{6} \\ $$$$\:{m}=\mathrm{8} \\ $$$${OR}, \\ $$$${The}\:{equation}\:{whose}\:{roots}\:{are} \\ $$$$−\mathrm{3},\:−\mathrm{1},\:\mathrm{1}\:\&\:\mathrm{3}\:{is} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{9}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$${Equating}\:{the}\:{coefficient}\:{ofx}^{\mathrm{2}} ,\:{we}\:{get} \\ $$$${m}+\mathrm{2}=\mathrm{10}\Rightarrow{m}=\mathrm{8} \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 08/Aug/20
V Nice!
$$\mathrm{V}\:\mathrm{Nice}! \\ $$
Commented by $@y@m last updated on 08/Aug/20
Thanks!
$${Thanks}! \\ $$

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