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 #