Question 104074


{{{x^2+2x-8=0}}} Start with the given equation



{{{x^2+2x=8}}} Add 8 to both sides



Take half of the x coefficient 2 to get 1 (ie {{{2/2=1}}})

Now square 1 to get 1 (ie {{{(1)^2=1}}})




{{{x^2+2x+1=8+1}}} Add this result (1) to both sides. Now the expression {{{x^2+2x+1}}} is a perfect square trinomial.





{{{(x+1)^2=8+1}}} Factor {{{x^2+2x+1}}} into {{{(x+1)^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+1)^2=9}}} Combine like terms on the right side


{{{x+1=0+-sqrt(9)}}} Take the square root of both sides


{{{x=-1+-sqrt(9)}}} Subtract 1 from both sides to isolate x.


So the expression breaks down to

{{{x=-1+sqrt(9)}}} or {{{x=-1-sqrt(9)}}}



{{{x=-1+3}}} or {{{x=-1-3}}}    Take the square root of 9 to get 3



{{{x=2}}} or {{{x=-4}}} Now combine like terms


So our answer is

{{{x=2}}} or {{{x=-4}}}



Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2+2x-8) }}} graph of {{{y=x^2+2x-8}}}


Here we can see that the x-intercepts are {{{x=2}}} and {{{x=-4}}}, so this verifies our answer.