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

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Answer 1 · 2016-04-06T12:44:04

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

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

$k^{2} + 6 k = 24$ $k^2 + 6k = 24$

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

$k^{2} + 2 \cdot k \cdot 3 + 3^{2} = 33$ $k^2 + 2 * k * 3 + 3^2 = 33$

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
${(k + 3)}^{2} = 33$ $(k + 3)^2 = 33$

$k + 3 = \sqrt{33}$ $k+ 3 = sqrt33$ or $k + 3 = - \sqrt{33}$ $k + 3 = -sqrt33$

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

How do you solve k^2 + 6k - 24 = 0k2+6k−24=0k^2 + 6k - 24 = 0 by completing the square?