^(@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

$\overset{@ bemath @}{}$ $What is m so the roots of x^{4} - (m + 2) x^{2} + 9 = 0$ $are in AP$
Commented by Her_Majesty last updated on 07/Aug/20

$s o t h e r o o t s o f . . . w h a t ? t h i s {d o e s n}^{'} t m a k e$ $s e n s e, {i t}^{'} s n o t a f u l l s e n t e n c e$
Commented by bemath last updated on 07/Aug/20

$yes . . you are right$
	Answered by john santu last updated on 07/Aug/20
	$^(◊JS⧫) say the roots are η , η+d, η+2d, η+3d By Vieta′s rule { ((4η+6d = 0 → d=−(2/3)η)),((η(η+d)(η+2d)(η+3d)=9)) :} ⇒η(η−(2/3)η)(η−(4/3)η)(η−(6/3)η)=9 η((1/3)η)(−(1/3)η)(−η)=9 η^4 =81 → { ((η=3)),((η=−3)) :} case(1) for η=3 substute to quartic equation ⇒3^4 −(m+2)3^2 +9=0 9−(m+2)+1=0 ⇒m=8 case(2) for η=−3 ⇒equal to η=3 therefore the value of m is 8.$
	$\overset{◊ JS ⧫}{}$ $say the roots are η, η + d, η + 2 d,$ $η + 3 d$ $By {Vieta}^{'} s rule$ ${\begin{cases} 4 η + 6 d = 0 \to d = - \frac{2}{3} η \\ η (η + d) (η + 2 d) (η + 3 d) = 9 \end{cases}$ $\Rightarrow η (η - \frac{2}{3} η) (η - \frac{4}{3} η) (η - \frac{6}{3} η) = 9$ $η (\frac{1}{3} η) (- \frac{1}{3} η) (- η) = 9$ $η^{4} = 81 \to {\begin{cases} η = 3 \\ η = - 3 \end{cases}$ $case (1) for η = 3$ $substute to quartic equation$ $\Rightarrow 3^{4} - (m + 2) 3^{2} + 9 = 0$ $9 - (m + 2) + 1 = 0 \Rightarrow m = 8$ $case (2) for η = - 3$ $\Rightarrow equal to η = 3$ $therefore the value of m is 8 .$
	Commented by Her_Majesty last updated on 07/Aug/20

	$h o w d o y o u k n o w w h a t i s a s k e d ?$ $c o f f e e c u p r e a d i n g ? t e l e p a t h y ?$
	Commented by john santu last updated on 07/Aug/20

	$by feeling$
	Answered by 1549442205PVT last updated on 07/Aug/20
	$The equation x^4 −(m+2)x^2 +9=0 has roots establish an AP if and only if two following conditions is satisfied:(set y=t^2 ) i)y^2 −(m+2)y^2 −(m+2)y+9=0(∗) has two positive roots y_1 ,y_2 .Since y_1 y_2 =9>0 ,this is equavalent to { ((Δ=(m+2)^2 −36>0)),((y_1 +y_2 =m+2>0)) :}⇔ { ((m^2 +4m−32>0)),((m+2>0)) :} ⇔ { ((((m−4)(m+8)>0)),((m+2>0)) :}⇔ { ((m∈(−∞;−8)∪(4;+∞))),((m>−2)) :} ⇔m∈(4;+∞)(∗∗) ii)Suppose that y_1 <y_2 and x_1 ,x_2 ,x_3 ,x_4 establish an arithmetic progression consecutively.Then we have: x_1 =−(√y_2 ) ,x_2 =−(√y_1 ) ,x_3 =(√y_1 ) ,x_4 =(√y_2 ) x_2 −x_1 =x_3 −x_2 =x_4 −x_3 =d ⇒ { ((2x_2 =x_1 +x_3 )),((2x_3 =x_2 +x_4 )) :}⇔ { ((4x_2 =2x_1 +2x_3 )),((2x_3 =x_2 +x_4 )) :} ⇒Adding up we get 4x_2 =2x_1 +x_2 +x_4 ⇔3x_2 =x_(1 ) (since x_1 +x_4 =x_2 +x_3 =0)(1) .On ther other hands, by Vieta′s theorem we have: y_1 +y_2 =m+2=x_2 ^2 +x_1 ^2 =x_3 ^2 +x_4 ^4 (2) y_1 y_2 =9=(x_1 x_2 )^2 =(x_3 x_4 )^2 ⇔x_1 x_2 =x_3 x_4 =3.From (1)we get 3x_2 ^2 =3⇒x_2 ^2 =1⇒x_2 =−1⇒x_1 =−3 Replace into (2) we obtain:m+2=10 ⇒m=8.This value satisfy the condition(∗∗) Consequently,The equation x^4 −(m+2)x^2 +9=0 has roots establish an arithmetic progression if an only if m=8$
	$The equation x^{4} - (m + 2) x^{2} + 9 = 0$ $has roots establish an AP$ $if and only if two following conditions$ $is satisfied : (set y = t^{2})$ $i) y^{2} - (m + 2) y^{2} - (m + 2) y + 9 = 0 ()$ $has two positive roots y_{1}, y_{2} . Since$ $y_{1} y_{2} = 9 > 0, this is equavalent to$ ${\begin{cases} Δ = {(m + 2)}^{2} - 36 > 0 \\ y_{1} + y_{2} = m + 2 > 0 \end{cases} \Leftrightarrow {\begin{cases} m^{2} + 4 m - 32 > 0 \\ m + 2 > 0 \end{cases}$ $\Leftrightarrow {\begin{cases} ((m - 4) (m + 8) > 0 \\ m + 2 > 0 \end{cases} \Leftrightarrow {\begin{cases} m \in (- \infty; - 8) \cup (4; + \infty) \\ m > - 2 \end{cases}$ $\Leftrightarrow m \in (4; + \infty) ( )$ $ii) Suppose that y_{1} < y_{2} and x_{1}, x_{2}, x_{3}, x_{4}$ $establish an arithmetic progression$ $consecutively . Then we have :$ $x_{1} = - \sqrt{y_{2}}, x_{2} = - \sqrt{y_{1}}, x_{3} = \sqrt{y_{1}}, x_{4} = \sqrt{y_{2}}$ $x_{2} - x_{1} = x_{3} - x_{2} = x_{4} - x_{3} = d$ $\Rightarrow {\begin{cases} {2 x}_{2} = x_{1} + x_{3} \\ {2 x}_{3} = x_{2} + x_{4} \end{cases} \Leftrightarrow {\begin{cases} {4 x}_{2} = {2 x}_{1} + {2 x}_{3} \\ {2 x}_{3} = x_{2} + x_{4} \end{cases}$ $\Rightarrow Adding up we get {4 x}_{2} = {2 x}_{1} + x_{2} + x_{4}$ $\Leftrightarrow {3 x}_{2} = x_{1} (since x_{1} + x_{4} = x_{2} + x_{3} = 0) (1)$ $. On ther other hands,$ $by {Vieta}^{'} s theorem we have :$ $y_{1} + y_{2} = m + 2 = x_{2}^{2} + x_{1}^{2} = x_{3}^{2} + x_{4}^{4} (2)$ $y_{1} y_{2} = 9 = {(x_{1} x_{2})}^{2} = {(x_{3} x_{4})}^{2}$ $\Leftrightarrow x_{1} x_{2} = x_{3} x_{4} = 3 . From (1) we get$ ${3 x}_{2}^{2} = 3 \Rightarrow x_{2}^{2} = 1 \Rightarrow x_{2} = - 1 \Rightarrow x_{1} = - 3$ $Replace into (2) we obtain : m + 2 = 10$ $\Rightarrow m = 8 . This value satisfy the condition ( *)$ $Consequently, The equation x^{4} - (m + 2) x^{2} + 9 = 0$ $has roots establish an arithmetic$ $progression if an only if m = 8$
	Answered by $@y@m last updated on 08/Aug/20
	$Let the roots be a−3d, a−d,a+d,a+3d Then, 4a=0 & (−3d)(−d)(d)(3d)=9 ⇒ a=0 & d=1 ∴ the roors are −3, −1, 1 & 3 Now, −(m+2)={(−3)+(−1)}.(1+3)+3+3 ⇒m+2=−16+6 m=8 OR, The equation whose roots are −3, −1, 1 & 3 is (x^2 −9)(x^2 −1)=0 ⇒x^4 −10x^2 +9=0 Equating the coefficient ofx^2 , we get m+2=10⇒m=8$
	$L e t t h e r o o t s b e$ $a - 3 d, a - d, a + d, a + 3 d$ $T h e n,$ $4 a = 0 & (- 3 d) (- d) (d) (3 d) = 9$ $\Rightarrow a = 0 & d = 1$ $∴ t h e r o o r s a r e$ $- 3, - 1, 1 & 3$ $N o w, - (m + 2) = {(- 3) + (- 1)} . (1 + 3) + 3 + 3$ $\Rightarrow m + 2 = - 16 + 6$ $m = 8$ $O R,$ $T h e e q u a t i o n w h o s e r o o t s a r e$ $- 3, - 1, 1 & 3 i s$ $(x^{2} - 9) (x^{2} - 1) = 0$ $\Rightarrow x^{4} - 10 x^{2} + 9 = 0$ $E q u a t i n g t h e c o e f f i c i e n t {o f x}^{2}, w e g e t$ $m + 2 = 10 \Rightarrow m = 8$
	Commented by Rasheed.Sindhi last updated on 08/Aug/20

	$V Nice!$
	Commented by $@y@m last updated on 08/Aug/20

	$T h a n k s!$
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