How do you solve the equation $x^{2} + 6 x + 13 = 0$ $x^2+6x+13=0$ by completing the square?

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How do you solve the equation $x^{2} + 6 x + 13 = 0$ $x^2+6x+13=0$ by completing the square?

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Answer 1 · 2017-01-08T18:38:07

$x = - 3 \pm 2 i$ $x = -3+-2i$

Explanation:

The difference of squares identity can be written:

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

Use this with $a = (x + 3)$ $a=(x+3)$ and $b = 2 i$ $b=2i$ as follows:

$0 = x^{2} + 6 x + 13$ $0 = x^2+6x+13$

$0 = x^{2} + 6 x + 9 + 4$ $color(white)(0) = x^2+6x+9+4$

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

$0 = ((x + 3) - 2 i) ((x + 3) + 2 i)$ $color(white)(0) = ((x+3)-2i)((x+3)+2i)$

$0 = (x + 3 - 2 i) (x + 3 + 2 i)$ $color(white)(0) = (x+3-2i)(x+3+2i)$

Hence:

$x = - 3 \pm 2 i$ $x = -3+-2i$

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How do you solve the equation $x^{2} + 6 x + 13 = 0$ $x^2+6x+13=0$ by completing the square?

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Explanation:

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