if the roots of the equation x^2 +(k+1)x+k=0 are α and β, find the value of the real constant k for which α=2β


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Question Number 210574 by ChantalYah last updated on 13/Aug/24

$i f t h e r o o t s o f t h e e q u a t i o n$ $x^{2} + (k + 1) x + k = 0$ $a r e α a n d β,$ $f i n d t h e v a l u e o f t h e$ $r e a l c o n s t a n t k f o r$ $w h i c h α = 2 β$
	Answered by mr W last updated on 13/Aug/24

	$α + β = - (k + 1) \Rightarrow 3 β = - (k + 1)$ $α β = k \Rightarrow 2 β^{2} = k$ $\Rightarrow 2 {(- \frac{k + 1}{3})}^{2} = k$ $\Rightarrow 2 k^{2} - 5 k + 2 = 0$ $\Rightarrow (k - 2) (2 k - 1) = 0$ $\Rightarrow k = 2, \frac{1}{2} ✓$
	Answered by Rasheed.Sindhi last updated on 13/Aug/24
	$Another way Roots are 2β and β { (((2β)^2 +(k+1)(2β)+k=0)),((β^2 +(k+1)β+k=0)) :} { ((4β^2 +2(k+1)β+k=0...(i))),((β^2 +(k+1)β+k=0...(ii))) :} (i)−(ii): 3β^2 +(k+1)β=0 3β+k+1=0 ; β≠0 β=−((k+1)/3) 4(ii)−(i): 2(k+1)β+3k=0 β=−((k+1)/3) ⇒2(k+1)(−((k+1)/3))+3k=0 −2(k+1)^2 +9k=0 2(k+1)^2 −9k=0 2k^2 −5k+2=0 (k−2)(2k−1)=0 k=2,(1/2)$
	$A n o t h e r w a y$ $R o o t s a r e 2 β a n d β$ ${\begin{cases} {(2 β)}^{2} + (k + 1) (2 β) + k = 0 \\ β^{2} + (k + 1) β + k = 0 \end{cases}$ ${\begin{cases} 4 β^{2} + 2 (k + 1) β + k = 0 . . . (i) \\ β^{2} + (k + 1) β + k = 0 . . . (i i) \end{cases}$ $(i) - (i i) : 3 β^{2} + (k + 1) β = 0$ $3 β + k + 1 = 0; β \neq 0$ $β = - \frac{k + 1}{3}$ $4 (i i) - (i) : 2 (k + 1) β + 3 k = 0$ $β = - \frac{k + 1}{3} \Rightarrow 2 (k + 1) (- \frac{k + 1}{3}) + 3 k = 0$ $- 2 {(k + 1)}^{2} + 9 k = 0$ $2 {(k + 1)}^{2} - 9 k = 0$ $2 k^{2} - 5 k + 2 = 0$ $(k - 2) (2 k - 1) = 0$ $k = 2, \frac{1}{2}$
	Answered by Rasheed.Sindhi last updated on 13/Aug/24

	$Still another way$ $x^{2} + (k + 1) x + k = 0$ $x^{2} + k x + x + k = 0$ $x (x + k) + (x + k) = 0$ $(x + k) (x + 1) = 0$ $x = - k \lor x = - 1$ $- k = 2 (- 1) \lor - 1 = 2 (- k)$ $k = 2 \lor k = \frac{1}{2}$
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