given-that-the-roots-of-the-equation-3x-2-4-2k-x-2k-0-are-and-find-the-value-of-k-for-which-3-

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

{\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 \dots (i) \\ 27 α^{2} - 3 (4 + 2 k) α + 2 k = 0 \dots (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