How do you solve $x^{2} - 4 x = 5$ $x^2 - 4x = 5$ by completing the square?

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Answer 1 · 2015-05-15T17:23:45

$x^{2} - 4 x = 5$ $x^2 - 4x = 5$
To write the Left Hand Side as a Perfect Square, we add 4 to both sides:

$x^{2} - 4 x + 4 = 5 + 4$ $x^2 - 4x + 4 = 5 + 4$

$x^{2} - 2.2 . x + 2^{2} = 9$ $x^2 -2.2.x + 2^2 = 9$
$x^{2} - 2.2 . x + 2^{2} = 3^{2}$ $x^2 -2.2.x + 2^2 = 3^2$

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
${(x - 2)}^{2} = 9$ $(x - 2)^2 = 9$

$x - 2 = \sqrt{9}$ $x - 2 = sqrt9$ or $x - 2 = - \sqrt{9}$ $x - 2 = -sqrt9$

$x = \sqrt{9} + 2$ $color(green)(x = sqrt 9 +2$ or $x = - \sqrt{9} + 2$ $color(green)(x = -sqrt 9 + 2$

How do you solve x2−4x=5x^2 - 4x = 5 by completing the square?