How do you solve $x^{2} + 8 x = 9$ $x^2+8x=9$ by completing the square?

Question

1294 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-26T14:39:19

The solutions are:
$x = 1, x = - 9$ $color(green)(x = 1 , x=-9$

$x^{2} + 8 x = 9$ $x^2 + 8x = 9$

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

$x^{2} + 8 x + 16 = 9 + 16$ $x^2 + 8x + 16 = 9+ 16$

$x^{2} + 2 \cdot x \cdot 4 + 4^{2} = 25$ $x^2 + 2* x * 4 + 4^2 = 25$

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,

${(x + 4)}^{2} = 25$ $(x+4)^2 = 25$

$x + 4 = \sqrt{25}$ $x + 4 = sqrt25$ or $x + 4 = - \sqrt{25}$ $x +4 = -sqrt25$

$x + 4 = 5$ $x + 4 = 5$ , $x + 4 = - 5$ $x +4 = -5$

$x = 5 - 4 = 1$ $x = 5- 4 =1$ ,

$x = - 5 - 4 = - 9$ $x = -5 -4 = -9$

The solutions are:
$x = 1, x = - 9$ $color(green)(x = 1 , x=-9$

How do you solve x2+8x=9x^2+8x=9 by completing the square?