Question Number 146611 by mathdanisur last updated on 14/Jul/21
$${y}={x}^{\mathrm{2}} +{x}\:,\:{y}=\mathrm{2}{x}^{\mathrm{2}} +{x}+{m}\:\:{and}\:\:{y}={kx}+{b} \\ $$$${if}\:{one}\:{point}\:{intersects},\:{find}\:\:\:{k}\centerdot{b}=? \\ $$
Answered by mr W last updated on 14/Jul/21
$${y}={x}^{\mathrm{2}} +{x}={kx}+{b} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{1}−{k}\right){x}−{b}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{1}−{k}\right)^{\mathrm{2}} +\mathrm{4}{b}=\mathrm{0} \\ $$$${y}=\mathrm{2}{x}^{\mathrm{2}} +{x}+{m}={kx}+{b} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\left(\mathrm{1}−{k}\right){x}+{m}−{b}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{1}−{k}\right)^{\mathrm{2}} +\mathrm{8}\left({b}−{m}\right)=\mathrm{0} \\ $$$$\mathrm{4}{b}=\mathrm{8}\left({b}−{m}\right) \\ $$$$\Rightarrow{b}=\mathrm{2}{m} \\ $$$$\left(\mathrm{1}−{k}\right)^{\mathrm{2}} =−\mathrm{8}{m} \\ $$$$\mathrm{1}−{k}=\pm\mathrm{2}\sqrt{−\mathrm{2}{m}} \\ $$$${k}=−\mathrm{1}\pm\mathrm{2}\sqrt{−\mathrm{2}{m}} \\ $$$$\Rightarrow{kb}=\mathrm{2}{m}\left(−\mathrm{1}\pm\mathrm{2}\sqrt{−\mathrm{2}{m}}\right) \\ $$
Commented by mathdanisur last updated on 14/Jul/21
$${Thankyou}\:{Ser},\:{cool} \\ $$