How do you solve using the completing the square method $x^{2} + 2 x + 4 = 0$ $x^2+2x+4=0$ ?

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How do you solve using the completing the square method $x^{2} + 2 x + 4 = 0$ $x^2+2x+4=0$ ?

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Answer 1 · 2017-04-05T06:15:16

$x = - 1 \pm \sqrt{3} i$ $x = -1+-sqrt(3)i$

Explanation:

Given:

$x^{2} + 2 x + 4 = 0$ $x^2+2x+4 = 0$

Completing the square we get:

$0 = x^{2} + 2 x + 4$ $0 = x^2+2x+4$

$0 = x^{2} + 2 x + 1 + 3$ $color(white)(0) = x^2+2x+1+3$

$0 = {(x + 1)}^{2} + 3$ $color(white)(0) = (x+1)^2+3$

Note that for any Real value of $x$ $x$ , we have:

${(x + 1)}^{2} \geq 0$ $(x+1)^2 >= 0$

and so:

${(x + 1)}^{2} + 3 \geq 3$ $(x+1)^2+3 >= 3$

To solve this quadratic we need to use Complex numbers.

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

We use this with $a = (x + 1)$ $a=(x+1)$ and $b = \sqrt{3} i$ $b=sqrt(3)i$ as follows:

$0 = {(x + 1)}^{2} + 3$ $0 = (x+1)^2+3$

$0 = {(x + 1)}^{2} - {(\sqrt{3} i)}^{2}$ $color(white)(0) = (x+1)^2-(sqrt(3)i)^2$

$0 = ((x + 1) - \sqrt{3} i) ((x + 1) + \sqrt{3} i)$ $color(white)(0) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)$

$0 = (x + 1 - \sqrt{3} i) (x + 1 + \sqrt{3} i)$ $color(white)(0) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)$

Hence:

$x = - 1 \pm \sqrt{3} i$ $x = -1+-sqrt(3)i$

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How do you solve using the completing the square method $x^{2} + 2 x + 4 = 0$ $x^2+2x+4=0$ ?

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