find k if Σ_(n=1) ^∞ (2k)(2)^(n−1) =64 hence find k if kx^2 +3x +4=0 has real roots.


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
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

	$. . . something seems to be missing . . .$ $x^{2} + \frac{3}{k} x + \frac{4}{k} = 0$ $x = \frac{3}{2 k} \pm \frac{\sqrt{9 - 16 k}}{2 k}$ $9 - 16 k ⩾ 0$ $k ⩽ \frac{9}{16}$
	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

	$still something wrong$ $\underset{n = 1}{\sum^{\infty}} (2 k) 2^{n - 1} = \infty \times sign (k)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum