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Question Number 210574 by ChantalYah last updated on 13/Aug/24
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
$${if}\:{the}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$${x}^{\mathrm{2}} +\left({k}+\mathrm{1}\right){x}+{k}=\mathrm{0} \\ $$$${are}\:\alpha\:{and}\:\beta, \\ $$$$\:{find}\:{the}\:{value}\:{of}\:{the} \\ $$$$\:{real}\:{constant}\:{k}\:{for} \\ $$$${which}\:\alpha=\mathrm{2}\beta \\ $$
Answered by mr W last updated on 13/Aug/24
α+β=−(k+1) ⇒3β=−(k+1) αβ=k ⇒2β^2 =k ⇒2(−((k+1)/3))^2 =k ⇒2k^2 −5k+2=0 ⇒(k−2)(2k−1)=0 ⇒k=2, (1/2) ✓
$$\alpha+\beta=−\left({k}+\mathrm{1}\right)\:\Rightarrow\mathrm{3}\beta=−\left({k}+\mathrm{1}\right) \\ $$$$\alpha\beta={k}\:\Rightarrow\mathrm{2}\beta^{\mathrm{2}} ={k} \\ $$$$\Rightarrow\mathrm{2}\left(−\frac{{k}+\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} ={k} \\ $$$$\Rightarrow\mathrm{2}{k}^{\mathrm{2}} −\mathrm{5}{k}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({k}−\mathrm{2}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{2},\:\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$
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)
$${Another}\:{way} \\ $$$${Roots}\:{are}\:\mathrm{2}\beta\:{and}\:\beta \\ $$$$\begin{cases}{\left(\mathrm{2}\beta\right)^{\mathrm{2}} +\left({k}+\mathrm{1}\right)\left(\mathrm{2}\beta\right)+{k}=\mathrm{0}}\\{\beta^{\mathrm{2}} +\left({k}+\mathrm{1}\right)\beta+{k}=\mathrm{0}}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{4}\beta^{\mathrm{2}} +\mathrm{2}\left({k}+\mathrm{1}\right)\beta+{k}=\mathrm{0}…\left({i}\right)}\\{\beta^{\mathrm{2}} +\left({k}+\mathrm{1}\right)\beta+{k}=\mathrm{0}…\left({ii}\right)}\end{cases}\: \\ $$$$\left({i}\right)−\left({ii}\right):\:\mathrm{3}\beta^{\mathrm{2}} +\left({k}+\mathrm{1}\right)\beta=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{3}\beta+{k}+\mathrm{1}=\mathrm{0}\:\:;\:\beta\neq\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\beta=−\frac{{k}+\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{4}\left({ii}\right)−\left({i}\right):\:\mathrm{2}\left({k}+\mathrm{1}\right)\beta+\mathrm{3}{k}=\mathrm{0} \\ $$$$\beta=−\frac{{k}+\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{2}\left({k}+\mathrm{1}\right)\left(−\frac{{k}+\mathrm{1}}{\mathrm{3}}\right)+\mathrm{3}{k}=\mathrm{0} \\ $$$$−\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{9}{k}=\mathrm{0} \\ $$$$\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9}{k}=\mathrm{0} \\ $$$$\mathrm{2}{k}^{\mathrm{2}} −\mathrm{5}{k}+\mathrm{2}=\mathrm{0} \\ $$$$\left({k}−\mathrm{2}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)=\mathrm{0} \\ $$$${k}=\mathrm{2},\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Aug/24
Still another way x^2 +(k+1)x+k=0 x^2 +kx+x+k=0 x(x+k)+(x+k)=0 (x+k)(x+1)=0 x=−k ∨ x=−1 −k=2(−1) ∨ −1=2(−k) k=2 ∨ k=(1/2)
$$\mathrm{Still}\:\mathrm{another}\:\mathrm{way} \\ $$$${x}^{\mathrm{2}} +\left({k}+\mathrm{1}\right){x}+{k}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{kx}+{x}+{k}=\mathrm{0} \\ $$$${x}\left({x}+{k}\right)+\left({x}+{k}\right)=\mathrm{0} \\ $$$$\left({x}+{k}\right)\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−{k}\:\vee\:{x}=−\mathrm{1} \\ $$$$−{k}=\mathrm{2}\left(−\mathrm{1}\right)\:\vee\:−\mathrm{1}=\mathrm{2}\left(−{k}\right) \\ $$$${k}=\mathrm{2}\:\vee\:{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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