find-k-if-n-1-2k-2-n-1-64-hence-find-k-if-kx-2-3x-4-0-has-real-roots-

Question Number 33185 by Rio Mike last updated on 12/Apr/18

f i n d k i f

\underset{n = 1}{\sum^{\infty}} (2 k) {(2)}^{n - 1} = 64

h e n c e f i n d k i f {k x}^{2} + 3 x + 4 = 0 h a s

r e a l r o o t s .

Answered by MJS last updated on 12/Apr/18

\dots something seems to be missing \dots

x^{2} + \frac{3}{k} x + \frac{4}{k} = 0

x = \frac{3}{2 k} \pm \frac{\sqrt{9 - 16 k}}{2 k}

9 - 16 k ⩾ 0

k ⩽ \frac{9}{16}

Commented by Rio Mike last updated on 12/Apr/18

y o u r r i g h t i c o r r e c t e d i t

Commented by MJS last updated on 12/Apr/18

still something wrong

\underset{n = 1}{\sum^{\infty}} (2 k) 2^{n - 1} = \infty \times sign (k)

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