High School

Which quadratic equation has roots [tex](-8+5i)[/tex] and [tex](-8-5i)[/tex]?

A. [tex]x^2 + 16x - 89 = 0[/tex]

B. [tex]x^2 - 16x + 89 = 0[/tex]

C. [tex]x^2 - 16x - 89 = 0[/tex]

D. [tex]x^2 + 16x + 89 = 0[/tex]

Answer :

To find the quadratic equation with the roots [tex]\(-8 + 5i\)[/tex] and [tex]\(-8 - 5i\)[/tex], we can use the fact that if a quadratic equation has roots [tex]\(a\)[/tex] and [tex]\(b\)[/tex], it can be expressed in the form:

[tex]\[
(x - a)(x - b) = 0
\][/tex]

Given the roots [tex]\(-8 + 5i\)[/tex] and [tex]\(-8 - 5i\)[/tex], let's substitute them into this form:

1. Substitute the roots into the factors:
[tex]\[
(x - (-8 + 5i))(x - (-8 - 5i)) = 0
\][/tex]

2. Simplify each term inside the parentheses:
[tex]\[
(x + 8 - 5i)(x + 8 + 5i) = 0
\][/tex]

3. Use the identity [tex]\((a + bi)(a - bi) = a^2 + b^2\)[/tex] to simplify the expression. Notice that this forms a difference of squares:
[tex]\[
(x + 8)^2 - (5i)^2
\][/tex]

4. Simplify each part:
- [tex]\((x + 8)^2 = x^2 + 16x + 64\)[/tex]
- [tex]\((5i)^2 = 25i^2 = -25\)[/tex] (since [tex]\(i^2 = -1\)[/tex])

5. Substitute back to get the quadratic expression:
[tex]\[
(x^2 + 16x + 64) - (-25) = x^2 + 16x + 64 + 25
\][/tex]

6. Combine the constant terms:
[tex]\[
x^2 + 16x + 89
\][/tex]

Therefore, the quadratic equation with the given roots is:

[tex]\[
x^2 + 16x + 89 = 0
\][/tex]