How do you solve $x^2-14x+19=0$ by completing the square?

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Answer 1 · 2016-04-18T17:15:06

The solutions are:
$color(green)(x = sqrt 30 + 7$ , $color(green)(x = -sqrt 30 + 7$

$x^2 - 14x + 19 =0$

$x^2 - 14x = - 19$

To write the Left Hand Side as a Perfect Square, we add 49 to both sides:

$x^2 - 14x + 49 = - 19 + 49$

$x^2 - 2 * x * 7 + 7^2 = 30$

Using the Identity $color(blue)((a-b)^2 = a^2 - 2ab + b^2$ , we get

$(x-7)^2 = 30$

$x - 7 = sqrt30$ or $x - 7 = -sqrt30$

$color(green)(x = sqrt 30 + 7$ or $color(green)(x = -sqrt 30 + 7$

How do you solve x^2-14x+19=0 x^2-14x+19=0 by completing the square?