Question 112076


{{{x^2+4x-4=0}}} Start with the given equation



{{{x^2+4x=4}}} Add 4 to both sides



Take half of the x coefficient 4 to get 2 (ie {{{4/2=2}}})

Now square 2 to get 4 (ie {{{(2)^2=4}}})




{{{x^2+4x+4=4+4}}} Add this result (4) to both sides. Now the expression {{{x^2+4x+4}}} is a perfect square trinomial.





{{{(x+2)^2=4+4}}} Factor {{{x^2+4x+4}}} into {{{(x+2)^2}}}  (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




{{{(x+2)^2=8}}} Combine like terms on the right side


{{{x+2=0+-sqrt(8)}}} Take the square root of both sides


{{{x=-2+-sqrt(8)}}} Subtract 2 from both sides to isolate x.


So the expression breaks down to

{{{x=-2+sqrt(8)}}} or {{{x=-2-sqrt(8)}}}



So our answer is approximately

{{{x=0.82842712474619}}} or {{{x=-4.82842712474619}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2+4x-4) }}} graph of {{{y=x^2+4x-4}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=0.82842712474619}}} and {{{x=-4.82842712474619}}}, so this verifies our answer.