Need help with a math questions. Ok so which quadratic equation has the roots 3+i and 3-i?

Question

1. x(squared)+6x-10=0 2. x(squared6x+8=0

3.x(squared)-6x+10=0 4. x(squared)-6x-8=0

Answer 1

x = 3 + i and x = 3 - i gives the equations:

x - 3 - i = 0 and x - 3 + i = 0.

If you multiply these together, you get the equation. Remember that i² = -1.

(x - 3 - i)(x - 3 + i) = 0.
x² - 6x + 9 - i² = 0
x² - 6x + 9 - (-1) = 0
x² - 6x + 10 = 0 <== this is your equation.

I hope that helps!! :-)

Answer 2

If your equation has roots r1 and r2 and the
equation is x² + ax + b = 0 then
sum of the roots = -a and product of the roots = b.
(You can get this from the quadratic formula.)
So a = -6
and b = (3+i)(3-i) = 10
Answer: x² -6x + 10 = 0.

Answer 3

You can figure this out by expanding the following equation: (x-(3+i))*(x-(3-i))=0. This equation essentially states that x equals 3+i and 3-i (the roots).

You could also use the quadratic formula and plug in the coefficients of each of the multiple choice answers to see which one produces the correct roots.

Good luck!

Answer 4

OK

x^2 + 6x - 10
-6 +- sqrt(36 - 4(1)(-10) / 2
-6 +- sqrt(36 +40)/ 2
(NOT this one - but what about 3)

x^2 -6x +10

6 +-sqrt(36 - 4(10)) /2
6 +-sqrt(-4) /2
6 +- 2i /2
3 +- i

Number 3 is the right one.

Hope that helps.

Answer 5

The 3rd choice: x^2 - 6x + 10 has 3+i and 3-i for roots.

that's it! :)