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Write a second degree polynomial with real coefficients and the given root; -2i

User Nstanard
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\text{For quadratic equations, if one root is complex, the other root will be its conjugate.}\\\\\text{So, the roots of the second degree polynomial are}~ \alpha = -2i ~~ \text{and}~~ \beta =2i \\\\\text{The equation is,}\\\\~~~~~~~x^2-(\alpha + \beta) x +\alpha \beta = 0\\ \\\implies x^2 -(-2i+2i)x +(2i)(-2i)=0\\\\\implies x^2 -0\cdot x -4i^2 =0\\ \\\implies x^2 -4(-1)=0\\\\\implies x^2 +4=0

User Medik
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Answer: x^2+4

Work Shown:

x = -2i

x^2 = (-2i)^2

x^2 = 4i^2

x^2 = 4(-1)

x^2 = -4

x^2+4 = 0

Side note: Since -2i is one root, this means 2i is the other conjugate root.

User Arctic Pi
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