How do you solve using the completing the square method $x^{2} + 8 x = 20$ $x^2+8x=20$ ?

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Answer 1 · 2016-03-09T14:00:24

The solutions are:
$x = 2$ $x = 2$ ,
$x = - 10$ $x = -10$

$x^{2} + 8 x = 20$ $x^2 +8x=20$

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

$x^{2} + 8 x + 16 = 20 + 16$ $x^2 + 8x + 16 = 20 + 16$

$x^{2} + 2 \cdot 4 \cdot x + 4^{2} = 36$ $x^2 + 2*4*x + 4^2 = 36$

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
${(x + 4)}^{2} = 36$ $(x+4)^2 = 36$

$x + 4 = \sqrt{36}$ $x + 4 = sqrt36$ or $x + 4 = - \sqrt{36}$ $x +4 = -sqrt36$

$x = 6 - 4$ $x = 6 - 4$ or $x = - 6 - 4$ $x = -6 -4$

$x = 2$ $x = 2$ or $x = - 10$ $x = -10$

The solutions are:
$x = 2$ $x = 2$ ,
$x = - 10$ $x = -10$

How do you solve using the completing the square method x2+8x=20x^2+8x=20?