Question Number 577 by Bek last updated on 31/Jan/15
![x^2 +2x+1=0](https://www.tinkutara.com/question/Q577.png)
$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$
Answered by Bek last updated on 31/Jan/15
![](https://www.tinkutara.com/question/Q578.png)
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Answered by prakash jain last updated on 31/Jan/15
![x^2 +2x+1=0 ⇒(x+1)^2 =0 ⇒x=−1](https://www.tinkutara.com/question/Q579.png)
$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{1} \\ $$
Answered by 123456 last updated on 31/Jan/15
![Δ=(2)^2 −4(1)(1)=4−4=0 x=((−(2)±(√0))/(2(1)))=−(2/2)=−1](https://www.tinkutara.com/question/Q580.png)
$$\Delta=\left(\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right)=\mathrm{4}−\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{−\left(\mathrm{2}\right)\pm\sqrt{\mathrm{0}}}{\mathrm{2}\left(\mathrm{1}\right)}=−\frac{\mathrm{2}}{\mathrm{2}}=−\mathrm{1} \\ $$