What is the quadratic equation that has the roots 3+i and 3−i ?(1 point)

Question

Responses

x2−6x+8=0
x squared minus 6 x plus 8 equals 0

x2+6x+10=0
x squared plus 6 x plus 10 equals 0

x2−6x+10=0
x squared minus 6 x plus 10 equals 0

x2+6x−10=0
x squared plus 6 x minus 10 equals 0

Answer 1

To find the quadratic equation with roots \( 3+i \) and \( 3-i \), we can use the fact that the quadratic can be expressed in the form:

\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]

where \( r_1 \) and \( r_2 \) are the roots.

1. Calculate the sum of the roots \( r_1 + r_2 \):

\[ (3+i) + (3-i) = 3 + i + 3 - i = 6 \]

2. Calculate the product of the roots \( r_1 \cdot r_2 \):

\[ (3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 9 + 1 = 10 \]

3. Substituting these values into the quadratic equation:

\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = x^2 - 6x + 10 = 0 \]

Thus, the quadratic equation is:

\[ x^2 - 6x + 10 = 0 \]

So, the correct response is:

x² - 6x + 10 = 0