Question Number 56384 by Gulay last updated on 15/Mar/19
Commented by Gulay last updated on 15/Mar/19
$$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$
Commented by mr W last updated on 16/Mar/19
$$\alpha,\beta={roots} \\ $$$$\alpha=\sqrt{\mathrm{10}}+\mathrm{4} \\ $$$$\alpha\beta=\mathrm{6} \\ $$$$\Rightarrow\beta=\frac{\mathrm{6}}{\:\sqrt{\mathrm{10}}+\mathrm{4}}=\frac{\mathrm{6}\left(\mathrm{4}−\sqrt{\mathrm{10}}\right)}{\mathrm{4}^{\mathrm{2}} −\mathrm{10}}=\mathrm{4}−\sqrt{\mathrm{10}} \\ $$$${C}=−\left(\alpha+\beta\right)=−\left(\sqrt{\mathrm{10}}+\mathrm{4}+\mathrm{4}−\sqrt{\mathrm{10}}\right)=−\mathrm{8} \\ $$
Answered by MJS last updated on 15/Mar/19
$${x}^{\mathrm{2}} +{cx}+\mathrm{6}=\mathrm{0} \\ $$$${x}=−\frac{{c}}{\mathrm{2}}\pm\frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{24}}}{\mathrm{2}} \\ $$$$−\frac{{c}}{\mathrm{2}}=\mathrm{4}\:\Rightarrow\:{c}=−\mathrm{8} \\ $$$$\frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{24}}}{\mathrm{2}}=\sqrt{\mathrm{10}}\:\Rightarrow\:{c}=\pm\mathrm{8} \\ $$$$\Rightarrow\:{c}=−\mathrm{8} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{4}−\sqrt{\mathrm{10}} \\ $$$$ \\ $$$$\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)=\left({x}−\mathrm{4}−\sqrt{\mathrm{10}}\right)\left({x}−\mathrm{4}+\sqrt{\mathrm{10}}\right)= \\ $$$$={x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{6} \\ $$