SOLUTION: Find the polynomial equation with real coefficients that has the given roots? 3 - i i think that this is a reverse on a quadratic equation, but i have no notes on the proc

Algebra ->  Radicals -> SOLUTION: Find the polynomial equation with real coefficients that has the given roots? 3 - i i think that this is a reverse on a quadratic equation, but i have no notes on the proc      Log On


   

Question 728367: Find the polynomial equation with real coefficients that has the given roots?
3 - i

i think that this is a reverse on a quadratic equation, but i have no notes on the procedure...can you please help
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x = 3 - i (given) or x = 3 + i (all complex roots come in conjugate pairs)

x - 3 = -i or x - 3 = i

(x - 3)^2 = (-i)^2 or (x - 3)^2 = i^2

(x - 3)^2 = i^2 or (x - 3)^2 = i^2

(x - 3)^2 = i^2

(x - 3)^2 = -1

(x - 3)^2 + 1 = 0 ----> Equation in Vertex Form: y+=+%28x-3%29%5E2+%2B+1
(x - 3)(x - 3) + 1 = 0

x^2 - 6x + 9 + 1 = 0

x^2 - 6x + 10 = 0 ----> Equation in Standard Form: y+=+x%5E2+-+6x+%2B+10