The roots of the equation 2x^2 +px+q=0 are 2α+β and α+2β. Find the values of p and q


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Question Number 145443 by 7770 last updated on 04/Jul/21

$T h e r o o t s o f t h e e q u a t i o n$ $2 x^{2} + p x + q = 0 a r e 2 α + β a n d$ $α + 2 β . F i n d t h e v a l u e s o f p a n d q$
	Answered by Olaf_Thorendsen last updated on 05/Jul/21

	$S = x_{1} + x_{2} = (2 α + β) + (α + 2 β) = 3 (α + β)$ $P = x_{1} x_{2} = (2 α + β) (α + 2 β) = 2 α^{2} + 5 α β + 2 β^{2}$ $S = - \frac{b}{a} = - \frac{p}{2} = 3 (α + β)$ $\Rightarrow p = - 6 (α + β)$ $P = \frac{c}{a} = \frac{q}{2} = 2 α^{2} + 5 α β + 2 β^{2}$ $\Rightarrow q = 4 α^{2} + 10 α β + 4 β^{2}$
	Answered by Rasheed.Sindhi last updated on 05/Jul/21
	$2x^2 +px+q=0; 2α+β &α+2β are roots._(−) 2(2α+β)^2 +p(2α+β)+q....(i) 2(α+2β)^2 +p(α+2β)+q....(ii) (i)−(ii): 2{(2α+β)^2 −(α+2β)^2 }+p{(2α+β)−(α+2β)}=0 −6α^2 +6β^2 −p(α−β)=0 p=((−6α^2 +6β^2 )/(α−β)))=((−6(α^2 −β^2 ))/(α−β))=−6(α+β) p=−6(α+β) (i)⇒ q=−2(2α+β)^2 −p(2α+β) =−2(2α+β)^2 −{−6(α+β)}(2α+β) =−8α^2 −8αβ−2β^2 +6(2α^2 +3αβ+β^2 ) q=4α^2 +10αβ+4β^2$
	$\underline{2 x^{2} + p x + q = 0; 2 α + β & α + 2 β a r e r o o t s .}$ $2 {(2 α + β)}^{2} + p (2 α + β) + q . . . . (i)$ $2 {(α + 2 β)}^{2} + p (α + 2 β) + q . . . . (i i)$ $(i) - (i i) :$ $2 {{(2 α + β)}^{2} - {(α + 2 β)}^{2}} + p {(2 α + β) - (α + 2 β)} = 0$ $- 6 α^{2} + 6 β^{2} - p (α - β) = 0$ $p = \frac{- 6 α^{2} + 6 β^{2}}{α - β)} = \frac{- 6 (α^{2} - β^{2})}{α - β} = - 6 (α + β)$ $p = - 6 (α + β)$ $(i) \Rightarrow q = - 2 {(2 α + β)}^{2} - p (2 α + β)$ $= - 2 {(2 α + β)}^{2} - {- 6 (α + β)} (2 α + β)$ $= - 8 α^{2} - 8 α β - 2 β^{2} + 6 (2 α^{2} + 3 α β + β^{2})$ $q = 4 α^{2} + 10 α β + 4 β^{2}$
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