Question 186580


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



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=0}}}, and {{{c=4}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(0) +- sqrt( (0)^2-4(1)(4) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=0}}}, and {{{c=4}}}



{{{x = (-0 +- sqrt( 0-4(1)(4) ))/(2(1))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (-0 +- sqrt( 0-16 ))/(2(1))}}} Multiply {{{4(1)(4)}}} to get {{{16}}}



{{{x = (-0 +- sqrt( -16 ))/(2(1))}}} Subtract {{{16}}} from {{{0}}} to get {{{-16}}}



{{{x = (-0 +- sqrt( -16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-0 +- 4*i)/(2)}}} Take the square root of {{{-16}}} to get {{{4*i}}}. 



{{{x = (-0 + 4*i)/(2)}}} or {{{x = (-0 - 4*i)/(2)}}} Break up the expression. 



{{{x = (-0)/(2) + (4*i)/(2)}}} or {{{x =  (-0)/(2) - (4*i)/(2)}}} Break up the fraction for each case. 



{{{x = 0+2*i}}} or {{{x =  0-2*i}}} Reduce. 



{{{x = 2*i}}} or {{{x = -2*i}}} Simplify. 



So the answers are {{{x = 2*i}}} or {{{x = -2*i}}} 

  

Note: these solutions are complex (ie non real)