Question

which of the following quadratic equations has roots of 2 + i and 2 - i? equation 1: $x^2 - 4x + 5 = 0$ equation 2: $x^2 + 4x + 5 = 0$ equation 3: $x^2 - 4x - 5 = 0$ (1 point) equation \boxed{}

Explanation:

Step1: Recall Vieta's formulas

For a quadratic equation \(ax^2 + bx + c = 0\) with roots \(r_1\) and \(r_2\), we have \(r_1 + r_2=-\frac{b}{a}\) and \(r_1r_2 = \frac{c}{a}\). Here, \(a = 1\) (since we can consider the general form \(x^2+bx + c=0\) as \(a = 1\)), \(r_1=2 + i\), \(r_2=2 - i\).

Step2: Calculate the sum of roots

\(r_1 + r_2=(2 + i)+(2 - i)=4\). From Vieta's formula, \(r_1 + r_2=-\frac{b}{a}\), and since \(a = 1\), we have \(-b=4\), so \(b=- 4\).

Step3: Calculate the product of roots

\(r_1r_2=(2 + i)(2 - i)=2^2 - i^2=4-(- 1)=5\). From Vieta's formula, \(r_1r_2=\frac{c}{a}\), and since \(a = 1\), \(c = 5\).

Step4: Form the quadratic equation

The quadratic equation with \(a = 1\), \(b=-4\), \(c = 5\) is \(x^2-4x + 5=0\), which is Equation 1.

Answer:

Equation 1: \(x^2 - 4x + 5 = 0\)