Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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!

$${Thanks}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com