if-the-roots-of-the-equation-x-2-k-1-x-k-0-are-and-find-the-value-of-the-real-constant-k-for-which-2-

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

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 \dots (i) \\ β^{2} + (k + 1) β + k = 0 \dots (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}