Question

What are the solutions if the equation x^2+4x+4=25

Answer 1

$\text{Use}\ (a+b)^2=a^2+2ab+b^2\\\\x^2+4x+4=25\\\\x^2+2(2)(x)+2^2=25\\\\(x+2)^2=25\iff x+2=\pm\sqrt{25}\\\\x+2=-5\ \vee\ x+2=5\qquad\text{subtract 2 from both sides}\\\\\boxed{x=-7\ \vee\ x=3}$

Other method:

$x^2+4x+4=25\qquad\text{subtracr 25 from both sides}\\\\x^2+4x-21=0\\\\x^2+7x-3x-21=0\\\\x(x+7)-3(x+7)=0\\\\(x+7)(x-3)=0\iff x+7=0\ \vee\ x-3=0\\\\x+7=0\qquad\text{subtract 7 from both sides}\\\boxed{x=-7}\\\\x-3=0\qquad\text{add 3 to both sides}\\\boxed{x=3}$