Question Number 33185 by Rio Mike last updated on 12/Apr/18
$${find}\:{k}\:{if} \\ $$$$\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}{k}\right)\left(\mathrm{2}\right)^{{n}−\mathrm{1}} =\mathrm{64} \\ $$$${hence}\:{find}\:{k}\:{if}\:\:{kx}^{\mathrm{2}} +\mathrm{3}{x}\:+\mathrm{4}=\mathrm{0}\:{has} \\ $$$${real}\:{roots}. \\ $$
Answered by MJS last updated on 12/Apr/18
$$...\mathrm{something}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{missing}... \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{3}}{{k}}{x}+\frac{\mathrm{4}}{{k}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}{k}}\pm\frac{\sqrt{\mathrm{9}−\mathrm{16}{k}}}{\mathrm{2}{k}} \\ $$$$\mathrm{9}−\mathrm{16}{k}\geqslant\mathrm{0} \\ $$$${k}\leqslant\frac{\mathrm{9}}{\mathrm{16}} \\ $$
Commented by Rio Mike last updated on 12/Apr/18
$${your}\:{right}\:{i}\:{corrected}\:{it} \\ $$
Commented by MJS last updated on 12/Apr/18
$$\mathrm{still}\:\mathrm{something}\:\mathrm{wrong} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}{k}\right)\mathrm{2}^{{n}−\mathrm{1}} =\infty×\mathrm{sign}\left({k}\right) \\ $$