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Question Number 210639 by ChantalYah last updated on 14/Aug/24
given that the roots of the equation 3x^2 −(4+2k)x+2k=0 are α and β find the value of k for which β=3α
$${given}\:{that}\:{the}\:{roots} \\ $$$$\:{of}\:{the}\:{equation} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\left(\mathrm{4}+\mathrm{2}{k}\right){x}+\mathrm{2}{k}=\mathrm{0} \\ $$$${are}\:\alpha\:{and}\:\beta \\ $$$${find}\:{the}\:{value}\:{of}\:{k} \\ $$$${for}\:{which}\:\beta=\mathrm{3}\alpha \\ $$
Answered by Rasheed.Sindhi last updated on 14/Aug/24
α+β=α+3α=4α=((4+2k)/3)⇒α^2 =(((2+k)/6))^2 αβ=α.3α=3α^2 =((2k)/3)⇒α^2 =((2k)/9) (((2+k)/6))^2 =((2k)/9) ((k^2 +4k+4)/4)=2k k^2 −4k+4=0 (k−2)^2 =0⇒k=2
$$\alpha+\beta=\alpha+\mathrm{3}\alpha=\mathrm{4}\alpha=\frac{\mathrm{4}+\mathrm{2}{k}}{\mathrm{3}}\Rightarrow\alpha^{\mathrm{2}} =\left(\frac{\mathrm{2}+{k}}{\mathrm{6}}\right)^{\mathrm{2}} \\ $$$$\alpha\beta=\alpha.\mathrm{3}\alpha=\mathrm{3}\alpha^{\mathrm{2}} =\frac{\mathrm{2}{k}}{\mathrm{3}}\Rightarrow\alpha^{\mathrm{2}} =\frac{\mathrm{2}{k}}{\mathrm{9}} \\ $$$$\left(\frac{\mathrm{2}+{k}}{\mathrm{6}}\right)^{\mathrm{2}} =\frac{\mathrm{2}{k}}{\mathrm{9}} \\ $$$$\frac{{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{4}}{\mathrm{4}}=\mathrm{2}{k} \\ $$$${k}^{\mathrm{2}} −\mathrm{4}{k}+\mathrm{4}=\mathrm{0} \\ $$$$\left({k}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{k}=\mathrm{2} \\ $$
Answered by mm1342 last updated on 14/Aug/24
3(x−α)(x−3α)=0 ⇒3x^2 −12αx+9α^2 =0 ⇒4+2k=12α & 2k=9α^2 ⇒9α^2 −12α+4=0 ⇒α=(2/3)⇒k=2 ✓
$$\mathrm{3}\left({x}−\alpha\right)\left({x}−\mathrm{3}\alpha\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}\alpha{x}+\mathrm{9}\alpha^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}+\mathrm{2}{k}=\mathrm{12}\alpha\:\:\&\:\:\mathrm{2}{k}=\mathrm{9}\alpha^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}\alpha^{\mathrm{2}} −\mathrm{12}\alpha+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\alpha=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow{k}=\mathrm{2}\:\checkmark\: \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 14/Aug/24
{ ((3α^2 −(4+2k)α+2k=0)),((3(3α)^2 −(4+2k)(3α)+2k=0)) :} { ((3α^2 −(4+2k)α+2k=0...(i))),((27α^2 −3(4+2k)α+2k=0...(ii))) :} (ii)−(i): 24α^2 −2(4+2k)α=0 12α−(4+2k)=0 ; α≠0 α=((2+k)/6) (ii)−9(i): 6(4+2k)α−16k=0 6(4+2k)(((2+k)/6))−16k=0 (2+k)^2 −8k=0 k^2 −4k+4=0 (k−2)^2 =0⇒k=2
$$\begin{cases}{\mathrm{3}\alpha^{\mathrm{2}} −\left(\mathrm{4}+\mathrm{2}{k}\right)\alpha+\mathrm{2}{k}=\mathrm{0}}\\{\mathrm{3}\left(\mathrm{3}\alpha\right)^{\mathrm{2}} −\left(\mathrm{4}+\mathrm{2}{k}\right)\left(\mathrm{3}\alpha\right)+\mathrm{2}{k}=\mathrm{0}}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{3}\alpha^{\mathrm{2}} −\left(\mathrm{4}+\mathrm{2}{k}\right)\alpha+\mathrm{2}{k}=\mathrm{0}…\left({i}\right)}\\{\mathrm{27}\alpha^{\mathrm{2}} −\mathrm{3}\left(\mathrm{4}+\mathrm{2}{k}\right)\alpha+\mathrm{2}{k}=\mathrm{0}…\left({ii}\right)}\end{cases}\: \\ $$$$\left({ii}\right)−\left({i}\right):\:\mathrm{24}\alpha^{\mathrm{2}} −\mathrm{2}\left(\mathrm{4}+\mathrm{2}{k}\right)\alpha=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{12}\alpha−\left(\mathrm{4}+\mathrm{2}{k}\right)=\mathrm{0}\:;\:\alpha\neq\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha=\frac{\mathrm{2}+{k}}{\mathrm{6}} \\ $$$$\left({ii}\right)−\mathrm{9}\left({i}\right):\:\mathrm{6}\left(\mathrm{4}+\mathrm{2}{k}\right)\alpha−\mathrm{16}{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{6}\left(\mathrm{4}+\mathrm{2}{k}\right)\left(\frac{\mathrm{2}+{k}}{\mathrm{6}}\right)−\mathrm{16}{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{2}+{k}\right)^{\mathrm{2}} −\mathrm{8}{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{k}^{\mathrm{2}} −\mathrm{4}{k}+\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\left({k}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{k}=\mathrm{2} \\ $$

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