What is the value of \(x\) in the equation \(9x^2 + 11 = 0\)?

Correct answer: \(\pm \dfrac{i\sqrt{11}}{3}\)

Solve: \(9x^2+11=0\Rightarrow 9x^2=-11\Rightarrow x^2=-\dfrac{11}{9}\). Taking square roots gives complex roots \(x=\pm\sqrt{-\dfrac{11}{9}}=\pm\dfrac{i\sqrt{11}}{3}\). Option C is incorrect because it gives real roots \(\pm\dfrac{\sqrt{11}}{3}\), which do not satisfy the equation; option D is wrong because \(x=0\) yields 11, not 0. Quick exam tip: check the sign of the discriminant or move the constant to the RHS — a negative value under the square root indicates imaginary (complex) roots involving \(i\).