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

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Answer 1 · 2016-04-18T17:15:06

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

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

$x^{2} - 14 x = - 19$ $x^2 - 14x = - 19$

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

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

$x^{2} - 2 \cdot x \cdot 7 + 7^{2} = 30$ $x^2 - 2 * x * 7 + 7^2 = 30$

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

${(x - 7)}^{2} = 30$ $(x-7)^2 = 30$

$x - 7 = \sqrt{30}$ $x - 7 = sqrt30$ or $x - 7 = - \sqrt{30}$ $x - 7 = -sqrt30$

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

How do you solve x2−14x+19=0x^2-14x+19=0 by completing the square?