Question 81001
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 = (-8 +- sqrt( 8^2-4*1*12 ))/(2*1) }}} 




==> {{{x = (-8 +- sqrt( 64-48))/(2) }}}




==>  {{{x = (-8 +- sqrt (16))/(2) }}}




==> {{{x = (-8 +- 4 )/(2) }}}




==> 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.