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


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Question Number 210639 by ChantalYah last updated on 14/Aug/24

$g i v e n t h a t t h e r o o t s$ $o f t h e e q u a t i o n$ $3 x^{2} - (4 + 2 k) x + 2 k = 0$ $a r e α a n d β$ $f i n d t h e v a l u e o f k$ $f o r w h i c h β = 3 α$
	Answered by Rasheed.Sindhi last updated on 14/Aug/24

	$α + β = α + 3 α = 4 α = \frac{4 + 2 k}{3} \Rightarrow α^{2} = {(\frac{2 + k}{6})}^{2}$ $α β = α . 3 α = 3 α^{2} = \frac{2 k}{3} \Rightarrow α^{2} = \frac{2 k}{9}$ ${(\frac{2 + k}{6})}^{2} = \frac{2 k}{9}$ $\frac{k^{2} + 4 k + 4}{4} = 2 k$ $k^{2} - 4 k + 4 = 0$ ${(k - 2)}^{2} = 0 \Rightarrow k = 2$
	Answered by mm1342 last updated on 14/Aug/24

	$3 (x - α) (x - 3 α) = 0$ $\Rightarrow 3 x^{2} - 12 α x + 9 α^{2} = 0$ $\Rightarrow 4 + 2 k = 12 α & 2 k = 9 α^{2}$ $\Rightarrow 9 α^{2} - 12 α + 4 = 0$ $\Rightarrow α = \frac{2}{3} \Rightarrow k = 2 ✓$
	Answered by Rasheed.Sindhi last updated on 14/Aug/24
	${ ((3α^2 −(4+2k)α+2k=0)),((3(3α)^2 −(4+2k)(3α)+2k=0)) :} { ((3α^2 −(4+2k)α+2k=0...(i))),((27α^2 −3(4+2k)α+2k=0...(ii))) :} (ii)−(i): 24α^2 −2(4+2k)α=0 12α−(4+2k)=0 ; α≠0 α=((2+k)/6) (ii)−9(i): 6(4+2k)α−16k=0 6(4+2k)(((2+k)/6))−16k=0 (2+k)^2 −8k=0 k^2 −4k+4=0 (k−2)^2 =0⇒k=2$
	${\begin{cases} 3 α^{2} - (4 + 2 k) α + 2 k = 0 \\ 3 {(3 α)}^{2} - (4 + 2 k) (3 α) + 2 k = 0 \end{cases}$ ${\begin{cases} 3 α^{2} - (4 + 2 k) α + 2 k = 0 . . . (i) \\ 27 α^{2} - 3 (4 + 2 k) α + 2 k = 0 . . . (i i) \end{cases}$ $(i i) - (i) : 24 α^{2} - 2 (4 + 2 k) α = 0$ $12 α - (4 + 2 k) = 0; α \neq 0$ $α = \frac{2 + k}{6}$ $(i i) - 9 (i) : 6 (4 + 2 k) α - 16 k = 0$ $6 (4 + 2 k) (\frac{2 + k}{6}) - 16 k = 0$ ${(2 + k)}^{2} - 8 k = 0$ $k^{2} - 4 k + 4 = 0$ ${(k - 2)}^{2} = 0 \Rightarrow k = 2$
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