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

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Answer 1 · 2017-01-31T04:52:13

$x=-1-isqrt3$ or $x=-1+isqrt3$

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

$ul(x^2+2xx x xx1 +1^2)-1^2+4=0$

and as $a^2+2ab+b^2=(a+b)^2$ , this is

$(x+1)^2-1+4=0$

or $(x+1)^2+3=0$

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

and using imaginary numbers $-3=(isqrt3)^2$ , as $i^2=-1$ and $(sqrt3)^2=3$

Hence, we can write $(x+1)^2+3=0$ as $(x+1)^2-(isqrt3)^2=0$

Using identity $a^2-b^2=(a+b)(a-b)$ , this becomes

$(x+1+isqrt3)(x+1-isqrt3)=0$

i.e. either $x=-1-isqrt3$ or $x=-1+isqrt3$ .

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