SOLUTION: please help me solve this? (x+4)^2=4

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: please help me solve this? (x+4)^2=4      Log On


   



Question 81001: please help me solve this?
(x+4)^2=4

Answer by praseenakos@yahoo.com(507) About Me  (Show Source):
You can put this solution on YOUR website!
QUESTION:
Solve,
(x+4)^2=4


ANSWER:


At first expand the expression using the identity (a+ b)^2 = a^2 + 2ab + b^2 as follows............


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

==> x^2 + 8x + 16 = 4


subtract 4 from both sides of this expression



==> x^2 + 8x + 16 - 4 = 4 - 4


==> x^2 + 8x + 12 = 0


this is in the form of a quadratic equation. we can solve this equation using quadratic formula.

Comaparing this equation with the stadard equation, ax^2 + bx + c =0, we have,



a = 1, b = 8 and c = 12


then the solution is given by,



x+=+%28-8+%2B-+sqrt%28+8%5E2-4%2A1%2A12+%29%29%2F%282%2A1%29+



==> x+=+%28-8+%2B-+sqrt%28+64-48%29%29%2F%282%29+



==> x+=+%28-8+%2B-+sqrt+%2816%29%29%2F%282%29+



==> x+=+%28-8+%2B-+4+%29%2F%282%29+



==> x = (-8 + 4)/2 or x = ( -8-4)/2




==> x = -4/2 or x = -12/2




==> x = -2 or x = -6




Hence the solution is x = -2 or -6



To check the solution, plugg any of these values in the given equation......then you can see that it satisfy the given equation



Hope you found this explanation useful.



Regards.



Praseena.