How do you solve $x^2 +8x +16=0$ ?

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Answer 1 · 2016-05-18T17:07:36

The solution is $x = -4$

$x^ 2 + 8x + 16 = 0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=1, b= 8, c=16$

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (8)^2-(4 * 1 * 16)$

$= 64 - 64 = 0$

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$x = (- 8+-sqrt(0))/(2 * 1 ) = (-8+-0 )/2$

$x = ( - 8 + 0 ) /2 = - 8/2 = - 4$

$x = ( - 8 - 0 ) /2 = - 8/2 = - 4$

How do you solve x^2 +8x +16=0x^2 +8x +16=0?