How to Find the Equation for a Quadratic with a Root of 1 + 3i?

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SUMMARY

The discussion centers on finding the quadratic equation given one root as 1 + 3i. It is established that if the equation has real coefficients, the other root must be the complex conjugate, 1 - 3i. The general form of the quadratic equation can be expressed as a(x - (1 + 3i))(x - (1 - 3i)) = 0, where 'a' is any non-zero constant. If real coefficients are not required, any complex number can serve as the second root.

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suganya
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Hi,

How to find the equation if one root of a quadratic equation is 1 + 3i.

Regards,
Suganya

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Does the equation has real co efficients? If it does, the other root is the complex conjugate of that root. If not, and you have the co efficients which I don't think you do, then just set (x - your root) as a factor and divide. If you don't know any of that information, no way to tell.
 
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Your "find the equation" implies that there is only one such equation. There isn't. If a quadratic equation has x0 and x1 as roots, then it must be of the form a(x-x0)(x- x1)= 0 where a can be any number.

You know that one root is 1+ 3i. As Gib Z said, if your equation must have real coefficients, then the other root must be its complex conjugate 1- 3i so any equation of the form a(x- 1- 3i)(x- 1+ 3i)= 0 will work. If you do not require real coefficients, choose any complex number at all for the second root!
 
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