text: How do you solve the quadratic with complex numbers given x^4+13x^2+36=0?
  
  Sep 28, 2016
  
  $x = \pm 2 i$ or $\pm 3 i$
  
  Explanation:
  
  ${x}^{4} + 13 {x}^{2} + 36 = 0$
  
  Let $z = {x}^{2} \to {z}^{2} + 13 z + 36 = 0$
  
  $\left(z + 4\right) \left(z + 9\right) = 0$
  
  Hence: $z = - 4$ or $- 9$
  
  $\therefore {x}^{2} = - 4$ or $- 9$
  
  $x = \sqrt{- 4}$ or $\sqrt{- 9}$
  
  $x = \pm 2 i$ or $\pm 3 i$
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