Question #b48d8

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Answer 1 · 2015-09-24T00:49:38

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

This quadratic expression should be solved by completing the square.

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

Minus seven from both sides,

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

Add four to both sides,

$x^{2} - 4 x + 4 = - 3$ $x^2-4x+4=-3$

Write the left hand side as a square,

${(x - 2)}^{2} = - 3$ $(x-2)^2=-3$

Take the square root of both sides,

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

Add two to both sides

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

or

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