If the roots of the equation 24x^4 −52x^3 +18x^2 +13x−6=0 are α , −α , β and (1/β). Find the value of α and β.


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Question Number 119396 by bemath last updated on 24/Oct/20

$I f t h e r o o t s o f t h e e q u a t i o n$ $24 x^{4} - 52 x^{3} + 18 x^{2} + 13 x - 6 = 0 a r e$ $α, - α, β a n d \frac{1}{β} . F i n d t h e v a l u e o f$ $α a n d β .$
Commented by Dwaipayan Shikari last updated on 24/Oct/20

$24 x^{4} - 52 x^{3} + 18 x^{2} + 13 x - 6 = 0$ $α - α + β + \frac{1}{β} = \frac{52}{24}$ $24 β^{2} - 52 β + 24 = 0 \Rightarrow 6 β^{2} - 13 β + 6 = 0$ $β = \frac{13 \pm \sqrt{169 - 144}}{12} = \frac{3}{2} o r \frac{2}{3}$ $α (- α) β . \frac{1}{β} = - \frac{6}{24}$ $α = \pm \frac{1}{2}$
	Answered by TANMAY PANACEA last updated on 24/Oct/20

	$α + (- α) + β + \frac{1}{β} = \frac{- (- 52)}{24} = \frac{13}{6}$ $6 β^{2} + 6 - 13 β = 0$ $6 β^{2} - 9 β - 4 β + 6 = 0$ $3 β (2 β - 3) - 2 (2 β - 3) = 0$ $(2 β - 3) (3 β - 2) = 0$ $β = \frac{2}{3} a n d \frac{3}{2}$ $α \times (- α) \times β \times \frac{1}{β} = \frac{- 6}{24}$ $α^{2} = \frac{1}{4} \to α = \pm \frac{1}{2}$
	Answered by $@y@m last updated on 24/Oct/20
	$24x^4 −52x^3 +18x^2 +13x−6=0 ≡ 24(x^2 −α^2 ){x^2 −(β+(1/β))x+1} ⇒24α^2 =6⇒α=±(1/2) Ans Part I Also, 24α^2 (β+(1/β))=13 β+(1/β)=((13)/6) 6β^2 −13β+6=0 β=((13±(√(169−144)))/(12)) β=((13±5)/(12))=((18)/(12)), (8/(12)) β=(3/2)′(2/3) Ans Part II$
	$24 x^{4} - 52 x^{3} + 18 x^{2} + 13 x - 6 = 0 \equiv$ $24 (x^{2} - α^{2}) {x^{2} - (β + \frac{1}{β}) x + 1}$ $\Rightarrow 24 α^{2} = 6 \Rightarrow α = \pm \frac{1}{2} A n s P a r t I$ $A l s o, 24 α^{2} (β + \frac{1}{β}) = 13$ $β + \frac{1}{β} = \frac{13}{6}$ $6 β^{2} - 13 β + 6 = 0$ $β = \frac{13 \pm \sqrt{169 - 144}}{12}$ $β = \frac{13 \pm 5}{12} = \frac{18}{12}, \frac{8}{12}$ $β = {\frac{3}{2}}^{'} \frac{2}{3} A n s P a r t II$
	Answered by 1549442205PVT last updated on 24/Oct/20
	$24x^4 −52x^3 +18x^2 +13x−6 =24(x+α)(x−α)(x−β)(x−(1/β)) =24(x^2 −α^2 )[x^2 −(β+(1/β))x+1] =24{x^4 −[α^2 (β+(1/β))+β+(1/β)]x^3 +(1−α^2 )x^2 +α^2 (β+(1/β))x−α^2 } ⇔ { ((−24α^2 (β+(1/β))=−52(1))),((24(1−α^2 )=18(2))),((24α^2 (β+(1/β))=13(3))),((−24α^2 =−6(4))) :} (1)⇒α^2 =1/4⇒α=±1/2.Replace (1) (3)we get β+(1/β)=((13)/6)⇔6β^2 −13β+6=0 Δ=13^2 −144=25⇒β=((13±5)/(12))∈{(3/2),(2/3)} Thus,α=±(1/2),β∈{(3/2),(2/3)}$
	${24 x}^{4} - {52 x}^{3} + {18 x}^{2} + 13 x - 6$ $= 24 (x + α) (x - α) (x - β) (x - \frac{1}{β})$ $= 24 (x^{2} - α^{2}) [x^{2} - (β + \frac{1}{β}) x + 1]$ $= 24 {x^{4} - [α^{2} (β + \frac{1}{β}) + β + \frac{1}{β}] x^{3}$ $+ (1 - α^{2}) x^{2} + α^{2} (β + \frac{1}{β}) x - α^{2}}$ $\Leftrightarrow {\begin{cases} - 24 α^{2} (β + \frac{1}{β}) = - 52 (1) \\ 24 (1 - α^{2}) = 18 (2) \\ 24 α^{2} (β + \frac{1}{β}) = 13 (3) \\ - 24 α^{2} = - 6 (4) \end{cases}$ $(1) \Rightarrow α^{2} = 1 / 4 \Rightarrow α = \pm 1 / 2 . Replace (1)$ $(3) we get β + \frac{1}{β} = \frac{13}{6} \Leftrightarrow 6 β^{2} - 13 β + 6 = 0$ $Δ = 13^{2} - 144 = 25 \Rightarrow β = \frac{13 \pm 5}{12} \in {\frac{3}{2}, \frac{2}{3}}$ $Thus, α = \pm \frac{1}{2}, β \in {\frac{3}{2}, \frac{2}{3}}$
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