# An equation has roots 1 - i and 1 + i. What is the equation?

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### 3 Answers

To solve questions of this type and remember it, the easiest way is to start by writing the statement of the question in math form, that is,

x = 1-i

and

x = 1+i

are two of the roots of the "unknown" equation. Then from them we rewrite each as a solution of the simplest connected equation possible, that is, write everything on the left side of the equal sign and the right side with of course zero, because it is a root of the equation:

From the first,

`x=1-irArr x-(1-i) = 0`

``From the second,

`x=1+irArr x-(1+i) =0`

Since we want both roots to belong to the same equation, just multiply the two and make them equal to zero:

`[x-(1-i)]*[x-(1+i)]=0`

`rArr x^2-x*(1+i)-(1-i)*x+(1-i)*(1+i)=0`

Then the two roots belong to the following second degree equation:

`x^2-2*x+2=0`

The equation with roots 1 - i and 1 + i has to be determined. The simplest equation with these roots is (x - (1 - i))(x - (1+i)) = 0.

(x - (1 - i))(x - (1+i)) = 0

=> (x - 1 + i)(x - 1 - i) = 0

=> (x - 1)^2 - i^2 = 0

=> x^2 + 1 - 2x + 1 = 0

=> x^2 - 2x + 2 = 0

**The simplest quadratic equation with the roots 1 + i and 1 - i is x^2 - 2x + 2 = 0**

1 - i and 1 + i

the first step is to group these solutions:

(x - 1 + i ) (x - 1 - i )

the reason we have x there is because in order to solve the equation would have looked something like this x - 1 + i =0 in order to get the solutions.

now F.O.I.L this stands for First Outer Inner Last it refers as to how you will multiply these numbers:

(x - 1 + i)(x - 1 - i) = 0

(x - 1) + i(x - 1) - i

(x - 1)^2 - i^2 = 0

i^2 is the same as -1

(x - 1)^2 -(-1) = 0

(x - 1) (x - 1)

FOIL the parenthesis, you should end up with:

x^2 -x -x +1

combine like terms

x^2 -2x + 1

add in the other result

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

x^2 - 2x + 2 = 0