Question Number 577 by Bek last updated on 31/Jan/15
$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$
Answered by Bek last updated on 31/Jan/15
$$ \\ $$
Answered by prakash jain last updated on 31/Jan/15
$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{1} \\ $$
Answered by 123456 last updated on 31/Jan/15
$$\Delta=\left(\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right)=\mathrm{4}−\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{−\left(\mathrm{2}\right)\pm\sqrt{\mathrm{0}}}{\mathrm{2}\left(\mathrm{1}\right)}=−\frac{\mathrm{2}}{\mathrm{2}}=−\mathrm{1} \\ $$