Question 100954


{{{x^2+4x-21=0}}} Start with the given equation



{{{x^2+4x=21}}} Add 21 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=21+4}}} Add this result (4) to both sides. Now the expression {{{x^2+4x+4}}} is a perfect square trinomial.





{{{(x+2)^2=21+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=25}}} Combine like terms on the right side


{{{x+2=0+-sqrt(25)}}} Take the square root of both sides


{{{x=-2+-sqrt(25)}}} Subtract 2 from both sides to isolate x.


So the expression breaks down to

{{{x=-2+sqrt(25)}}} or {{{x=-2-sqrt(25)}}}



{{{x=-2+5}}} or {{{x=-2-5}}}    Take the square root of 25 to get 5



{{{x=3}}} or {{{x=-7}}} Now combine like terms


So our answer is

{{{x=3}}} or {{{x=-7}}}



Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2+4x-21) }}} graph of {{{y=x^2+4x-21}}}


Here we can see that the x-intercepts are {{{x=3}}} and {{{x=-7}}}, so this verifies our answer.