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

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Answer 1 · 2015-03-30T12:37:01

Since ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2 = x^2+2ax + a^2$

If $x^{2} + 8 x$ $x^2 +8x$ are the first 2 terms of a squared expression
the third term must be ${(\frac{8}{2})}^{2} = 16$ $(8/2)^2 = 16$

Add $16$ $16$ to each side of the equation, giving
$x^{2} + 8 x + 16 = 23$ $x^2+8x+16 = 23$
which is the same as
${(x + 4)}^{2} = 23$ $(x+4)^2 = 23$

Taking the square root of both sides
$x + 4 = \pm \sqrt{23}$ $x+4 = +-sqrt(23)$

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

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