Sequoyah
Sequoyah Sequoyah
  • 01-11-2017
  • Mathematics
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What is the solution of the equation when solved over the complex numbers?

x^2+24=0
x =
or x =

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budwilkins
budwilkins budwilkins
  • 01-11-2017

[tex] {x}^{2} + 24 = 0 \\ (x - \sqrt{24})(x + \sqrt{24} ) = 0 \\ (x - \sqrt{4} \sqrt{6})(x + \sqrt{4} \sqrt{6}) = 0 [/tex]
[tex](x - 2 \sqrt{6})(x + 2 \sqrt{6} ) = 0 \\ so \: x = 2 \sqrt{6} = 4.899 \: and \\ x = - 2 \sqrt{6} = - 4.899[/tex]
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