How do you solve using completing the square method $x^{2} - 6 x - 7 = 0$ $x^2-6x-7=0$ ?

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Answer 1 · 2016-05-03T11:56:08

The solutions are:
$x = 7$ $color(green)(x = 7$ , $x = - 1$ $color(green)(x = -1$

$x^{2} - 6 x - 7 = 0$ $x^2 - 6x - 7 =0$

$x^{2} - 6 x = 7$ $x^2 - 6x = 7$

To write the Left Hand Side as a Perfect Square, we add 9 to both sides:

$x^{2} - 6 x + 9 = 7 + 9$ $x^2 - 6x + color(blue)(9) = 7 + color(blue)(9$

$x^{2} - 2 \cdot x \cdot 3 + 3^{2} = 16$ $x^2 - 2 * x * 3 + 3^2 = 16$

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

${(x - 3)}^{2} = 16$ $(x- 3 )^2 = 16$

$x - 3 = \sqrt{16}$ $x - 3 = sqrt16$ or $x - 3 = - \sqrt{16}$ $x - 3 = -sqrt16$

$x - 3 = 4$ $x - 3 = 4$ or $x - 3 = - 4$ $x - 3 = -4$

$x = 4 + 3$ $x = 4 + 3$ or $x = - 4 + 3$ $x = -4 + 3$

$x = 7$ $color(green)(x = 7$ , $x = - 1$ $color(green)(x = -1$

How do you solve using completing the square method x^2-6x-7=0x2−6x−7=0x^2-6x-7=0?