How do you solve $k^2 + 6k - 24 = 0$ by completing the square?

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Answer 1 · 2016-04-06T12:44:04

The solutions are:
$color(green)(k = sqrt 33 - 3$ or , $color(green)(k = -sqrt 33 -3$

$k^2 + 6k - 24 = 0$

$k^2 + 6k = 24$

To write the Left Hand Side as a Perfect Square, we add 9 to both sides
$k^2 + 6k + 9 = 24 + 9$

$k^2 + 2 * k * 3 + 3^2 = 33$

Using the Identity $color(blue)((a+b)^2 = a^2 + 2ab + b^2$ , we get
$(k + 3)^2 = 33$

$k+ 3 = sqrt33$ or $k + 3 = -sqrt33$

$color(green)(k = sqrt 33 - 3$ or $color(green)(x = -sqrt 33 -3$

How do you solve k^2 + 6k - 24 = 0k^2 + 6k - 24 = 0 by completing the square?