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

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

$${find}\:{k}\:{if} \\ $$$$\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}{k}\right)\left(\mathrm{2}\right)^{{n}−\mathrm{1}} =\mathrm{64} \\ $$$${hence}\:{find}\:{k}\:{if}\:\:{kx}^{\mathrm{2}} +\mathrm{3}{x}\:+\mathrm{4}=\mathrm{0}\:{has} \\ $$$${real}\:{roots}. \\ $$

Answered by MJS last updated on 12/Apr/18

...something seems to be missing... x^2 +(3/k)x+(4/k)=0 x=(3/(2k))±((√(9−16k))/(2k)) 9−16k≥0 k≤(9/(16))

$$…\mathrm{something}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{missing}… \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{3}}{{k}}{x}+\frac{\mathrm{4}}{{k}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}{k}}\pm\frac{\sqrt{\mathrm{9}−\mathrm{16}{k}}}{\mathrm{2}{k}} \\ $$$$\mathrm{9}−\mathrm{16}{k}\geqslant\mathrm{0} \\ $$$${k}\leqslant\frac{\mathrm{9}}{\mathrm{16}} \\ $$

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

$${your}\:{right}\:{i}\:{corrected}\:{it} \\ $$

Commented by MJS last updated on 12/Apr/18

still something wrong Σ_(n=1) ^∞ (2k)2^(n−1) =∞×sign(k)

$$\mathrm{still}\:\mathrm{something}\:\mathrm{wrong} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}{k}\right)\mathrm{2}^{{n}−\mathrm{1}} =\infty×\mathrm{sign}\left({k}\right) \\ $$

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