# Finding original algebraic equation

1. Oct 7, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data

if I have a root of a +/- bi

i = imaginary number

what should the original algebra eqution?

2. Relevant equations

3. The attempt at a solution

2. Oct 7, 2009

### Staff: Mentor

The original polynomial should have factors of (x - (a + bi))(x - (a - bi)), or equivalently, a single factor of (x2 - 2ax + a2 + b2). If the degree of the polynomial you're working with is larger than 2, there will be other roots, as well.

3. Oct 7, 2009

### HallsofIvy

x- (a+ bi) and x- (a- bi) are factors. Multiply $(x- (a+bi))(x- (a- bi)= ((x-a)- bi)((x-a)+bi)= (x-a)^2+ b^2$ and set it equal to 0.

4. Oct 7, 2009

### -EquinoX-

thanks Mark, shouldn't it be a^2 - b^2?

5. Oct 8, 2009

### Staff: Mentor

Nope. The a2 + b2 comes from multiplying (-(a + bi))(-(a - bi)) = (a + bi)(a - bi) = a2 -b2i2 = a2 + b2.

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