Question 119189
Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{9*x^2+9=0}}} (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like {{{9*x^2+0*x+9=0}}}  notice {{{a=9}}}, {{{b=0}}}, and {{{c=9}}})





{{{x = (0 +- sqrt( (0)^2-4*9*9 ))/(2*9)}}} Plug in a=9, b=0, and c=9




{{{x = (0 +- sqrt( 0-4*9*9 ))/(2*9)}}} Square 0 to get 0  




{{{x = (0 +- sqrt( 0+-324 ))/(2*9)}}} Multiply {{{-4*9*9}}} to get {{{-324}}}




{{{x = (0 +- sqrt( -324 ))/(2*9)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (0 +- 18*i)/(2*9)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (0 +- 18*i)/(18)}}} Multiply 2 and 9 to get 18




After simplifying, the quadratic has roots of



{{{x=i}}} or {{{x= - i}}}



So the answer is B)