How do you factor and solve #x^2+4x+4=0 #?

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Answer 1 · 2016-09-19T21:46:56

#x = - 2#

We have: #x^(2) + 4 x + 4 = 0#

Let's factorise this equation using the middle-term break:

#=> x^(2) + 2 x + 2 x + 4 = 0#

#=> x (x + 2) + 2 (x + 2) = 0#

#=> (x + 2) (x + 2) = 0#

#=> (x + 2)^(2) = 0#

#=> x + 2 = 0#

#=> x = - 2#

Therefore, the solution to the equation is #x = - 2#.