How do you solve using the completing the square method $x^{2} - 8 x - 3 = 0$ $x^2 - 8x - 3 = 0$ ?

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Answer 1 · 2016-03-23T17:09:45

The solutions are:
$x = \sqrt{19} + 4$ $color(green)(x = sqrt 19 + 4$
$x = - \sqrt{19} + 4$ $color(green)(x = -sqrt 19 + 4$

$x^{2} - 8 x - 3 = 0$ $x^2 - 8x - 3 = 0$

$x^{2} - 8 x = 3$ $x^2 - 8x = 3$

To write the Left Hand Side as a Perfect Square, we add 16 to both sides:

$x^{2} - 8 x + 16 = 3 + 16$ $x^2 - 8x + 16 = 3 + 16$

$x^{2} - 2 \cdot x \cdot 4 + 4^{2} = 19$ $x^2 - 2 * x * 4 + 4^2 = 19$

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:

${(x - 4)}^{2} = 19$ $(x-4)^2 = 19$

$x - 4 = \sqrt{19}$ $x - 4 = sqrt19$ or $x - 4 = - \sqrt{19}$ $x -4 = -sqrt19$

$x = \sqrt{19} + 4$ $color(green)(x = sqrt 19 + 4$ or $x = - \sqrt{19} + 4$ $color(green)(x = -sqrt 19 + 4$

How do you solve using the completing the square method x2−8x−3=0x^2 - 8x - 3 = 0?