How do you solve $x^{2} + 8 x = 7$ $x^2 +8x = 7$ by completing the square?

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Answer 1 · 2016-04-04T16:14:41

The solutions are:
$x = \sqrt{23} - 4$ $color(green)(x = sqrt 23 - 4$ or $x = - \sqrt{23} - 4$ $color(green)(x = -sqrt 23 - 4$

$x^{2} + 8 x = 7$ $x^2 + 8x = 7$

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

$x^{2} + 8 x + 16 = 7 + 16$ $x^2 + 8x + 16 = 7 + 16$

$x^{2} + 2 \cdot x \cdot 4 + 4^{2} = 23$ $x^2 + 2 * x * 4 + 4^2 = 23$

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

${(x + 4)}^{2} = 23$ $(x+4)^2 = 23$

$x + 4 = \sqrt{23}$ $x + 4 = sqrt23$ or $x + 4 = - \sqrt{23}$ $x +4 = -sqrt23$

$x = \sqrt{23} - 4$ $color(green)(x = sqrt 23 - 4$ or $x = - \sqrt{23} - 4$ $color(green)(x = -sqrt 23 - 4$

How do you solve x2+8x=7x^2 +8x = 7 by completing the square?