How do you solve $x^{2} - 4 x + 7 = 0$ $x^2 - 4x + 7 =0$ by completing the square?

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How do you solve $x^{2} - 4 x + 7 = 0$ $x^2 - 4x + 7 =0$ by completing the square?

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Answer 1 · 2017-08-10T06:16:04

$x = 2 \pm \sqrt{3} i$ $x = 2+-sqrt(3)i$

Explanation:

Given:

$x^{2} - 4 x + 7 = 0$ $x^2-4x+7=0$

While completing the square we will find that this takes the form of the sum of a square and a positive number. As a result it has no solution in real numbers, but we can solve it using complex numbers.

The imaginary unit $i$ $i$ satisfies $i^{2} = - 1$ $i^2=-1$

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 can use this with $a = (x - 2)$ $a=(x-2)$ and $b = \sqrt{3} i$ $b=sqrt(3)i$ as follows:

$0 = x^{2} - 4 x + 7$ $0 = x^2-4x+7$

$0 = x^{2} - 4 x + 4 + 3$ $color(white)(0) = x^2-4x+4+3$

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

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

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

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

Hence the two roots are:

$x = 2 + \sqrt{3} i$ $x = 2+sqrt(3)i" "$ and $x = 2 - \sqrt{3} i$ $" "x = 2-sqrt(3)i$

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How do you solve $x^{2} - 4 x + 7 = 0$ $x^2 - 4x + 7 =0$ by completing the square?

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How do you solve $x^{2} - 4 x + 7 = 0$ $x^2 - 4x + 7 =0$ by completing the square?