Question

Which of the following quadratic equations has roots of 2+i
and 2−i
?

Equation 1: x2−4x+5=0

Equation 2: x2+4x+5=0

Equation 3: x2−4x−5=0

Answer 1

To find which quadratic equation has roots \(2+i\) and \(2-i\), we can use the fact that a quadratic equation can be expressed based on its roots. If the roots of a quadratic equation are \(r_1\) and \(r_2\), the quadratic equation can be written as:

\[
(x - r_1)(x - r_2) = 0
\]

In this case, the roots given are \(r_1 = 2 + i\) and \(r_2 = 2 - i\). Therefore, we can write:

\[
(x - (2 + i))(x - (2 - i)) = 0
\]

Now we can simplify this expression:

\[
(x - (2 + i))(x - (2 - i)) = (x - 2 - i)(x - 2 + i)
\]

This is a difference of squares:

\[
= ((x - 2) - i)((x - 2) + i) = (x - 2)^2 - i^2
\]

Since \(i^2 = -1\), we have:

\[
= (x - 2)^2 + 1
\]

Now we expand \((x - 2)^2 + 1\):

\[
= (x^2 - 4x + 4) + 1 = x^2 - 4x + 5
\]

We find that the quadratic equation with roots \(2+i\) and \(2-i\) is:

\[
x^2 - 4x + 5 = 0
\]

Now we can compare with the given equations:

1. Equation 1: \(x^2 - 4x + 5 = 0\) (matches)
2. Equation 2: \(x^2 + 4x + 5 = 0\) (does not match)
3. Equation 3: \(x^2 - 4x - 5 = 0\) (does not match)

Therefore, the answer is **Equation 1: \(x^2 - 4x + 5 = 0\)**.