# How do you solve x^2+2x+4=0 by completing the square?

Jan 31, 2017

$x = - 1 - i \sqrt{3}$ or $x = - 1 + i \sqrt{3}$

#### Explanation:

${x}^{2} + 2 x + 4 = 0$ can be written as

$\underline{{x}^{2} + 2 \times x \times 1 + {1}^{2}} - {1}^{2} + 4 = 0$

and as ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$, this is

${\left(x + 1\right)}^{2} - 1 + 4 = 0$

or ${\left(x + 1\right)}^{2} + 3 = 0$

Now, to solve the equation, we must convert it to form ${a}^{2} - {b}^{2}$, but as we have $+ 3$, we write it as $- \left(- 3\right)$

and using imaginary numbers $- 3 = {\left(i \sqrt{3}\right)}^{2}$, as ${i}^{2} = - 1$ and ${\left(\sqrt{3}\right)}^{2} = 3$

Hence, we can write ${\left(x + 1\right)}^{2} + 3 = 0$ as ${\left(x + 1\right)}^{2} - {\left(i \sqrt{3}\right)}^{2} = 0$

Using identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this becomes

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

i.e. either $x = - 1 - i \sqrt{3}$ or $x = - 1 + i \sqrt{3}$.