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How can you solve the equation x2 plus 3i equals 0? - Answers

x2+3i=0 so x2=-3i x=square root of (-3i)=square root (-3)square root (i) =i(square root(3)([1/(square root (2)](1+i) and i(square root(3)([-1/(square root (2)](1+i) You can multiply through by i if you want, but I left it since it shows you where the answer came from. Note: The square root of i is 1/square root 2(1+i) and -1/square root of 2 (1+i) to see this, try and square them!



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How can you solve the equation x2 plus 3i equals 0? - Answers

https://math.answers.com/calculus/How_can_you_solve_the_equation_x2_plus_3i_equals_0

x2+3i=0 so x2=-3i x=square root of (-3i)=square root (-3)square root (i) =i(square root(3)([1/(square root (2)](1+i) and i(square root(3)([-1/(square root (2)](1+i) You can multiply through by i if you want, but I left it since it shows you where the answer came from. Note: The square root of i is 1/square root 2(1+i) and -1/square root of 2 (1+i) to see this, try and square them!



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https://math.answers.com/calculus/How_can_you_solve_the_equation_x2_plus_3i_equals_0

How can you solve the equation x2 plus 3i equals 0? - Answers

x2+3i=0 so x2=-3i x=square root of (-3i)=square root (-3)square root (i) =i(square root(3)([1/(square root (2)](1+i) and i(square root(3)([-1/(square root (2)](1+i) You can multiply through by i if you want, but I left it since it shows you where the answer came from. Note: The square root of i is 1/square root 2(1+i) and -1/square root of 2 (1+i) to see this, try and square them!

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      x2+3i=0 so x2=-3i x=square root of (-3i)=square root (-3)square root (i) =i(square root(3)([1/(square root (2)](1+i) and i(square root(3)([-1/(square root (2)](1+i) You can multiply through by i if you want, but I left it since it shows you where the answer came from. Note: The square root of i is 1/square root 2(1+i) and -1/square root of 2 (1+i) to see this, try and square them!

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