How do you solve using the completing the square method $x^{2} - 4 x - 9 = 0$ $x^2 - 4x - 9 = 0$ ?

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Answer 1 · 2016-03-30T16:46:55

The solutions are:
$x = \sqrt{13} + 2$ $color(green)(x = sqrt 13 + 2$ or $x = - \sqrt{13} + 2$ $color(green)(x = -sqrt 13 + 2$

$x^{2} - 4 x - 9 = 0$ $x^2 - 4x - 9 =0$

$x^{2} - 4 x = 9$ $x^2 - 4x = 9$

To write the Left Hand Side as a Perfect Square, we add 4 to both sides:

$x^{2} - 4 x + 4 = 9 + 4$ $x^2 - 4x+ 4 = 9 + 4$

$x^{2} - 2 \cdot x \cdot 2 + 2^{2} = 13$ $x^2 - 2* x * 2 + 2^2 = 13$

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

${(x - 2)}^{2} = 13$ $(x-2)^2 = 13$

$x - 2 = \sqrt{13}$ $x - 2 = sqrt13$ or $x - 2 = - \sqrt{13}$ $x -2 = -sqrt13$

$x = \sqrt{13} + 2$ $color(green)(x = sqrt 13 + 2$ or $x = - \sqrt{13} + 2$ $color(green)(x = -sqrt 13 + 2$

How do you solve using the completing the square method x2−4x−9=0x^2 - 4x - 9 = 0?