Solve the following
Question:

$x^{2}+x+1=0$

 

Solution:

We have:

$x^{2}+x+1=0$

$\Rightarrow x^{2}+x+\frac{1}{4}+\frac{3}{4}=0$

$\Rightarrow x^{2}+\left(\frac{1}{2}\right)^{2}+2 \times x \times \frac{1}{2}-\left(\frac{\sqrt{3} i}{2}\right)^{2}=0$

$\Rightarrow\left(x+\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3} i}{2}\right)^{2}=0$

$\Rightarrow\left(x+\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\left(x+\frac{1}{2}-\frac{\sqrt{3} i}{2}\right)=0$

$\Rightarrow\left(x+\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)=0$ or, $\left(x+\frac{1}{2}-\frac{\sqrt{3} i}{2}\right)=0$

$\Rightarrow x=-\frac{1}{2}-\frac{\sqrt{3} i}{2}$ or, $x=-\frac{1}{2}+\frac{\sqrt{3} i}{2}$

Hence, the roots of the equation are $-\frac{1}{2}-i \frac{\sqrt{3}}{2}$ and $-\frac{1}{2}+i \frac{\sqrt{3}}{2}$.

Administrator

Leave a comment

Please enter comment.
Please enter your name.