Which equation has the complex number 4-3i as a root?

1) x^2+6x-25=0
2) x^2-6x+25=0
3) x^2+8x-25=0
4) x^2-8x+25=0

Can anyone please teach me how to do it?THANKS A LOT!!!

To find which equation has the complex number 4-3i as a root, we need to use the fact that complex roots occur in conjugate pairs.

First, let's rewrite the complex number 4-3i using the fact that complex conjugates have the same real part but opposite signs on the imaginary part.

The conjugate of 4-3i is 4+3i.

Now, to form the equation with the given complex number as a root, we can use the fact that if a+bi is a root of a quadratic equation, then (x - a - bi) and (x - a + bi) are factors of the equation.

Let's check each equation using the given complex number as a root:

1) x^2 + 6x - 25 = 0
The roots of this equation are not 4-3i and its conjugate 4+3i.

2) x^2 - 6x + 25 = 0
The roots of this equation are not 4-3i and its conjugate 4+3i.

3) x^2 + 8x - 25 = 0
The roots of this equation are not 4-3i and its conjugate 4+3i.

4) x^2 - 8x + 25 = 0
The roots of this equation are indeed 4-3i and its conjugate 4+3i.

Observing that only equation 4) has the complex number 4-3i as a root, we can conclude that the correct equation is x^2 - 8x + 25 = 0.

To determine which equation has the complex number 4-3i as a root, we need to substitute the given complex number into each equation and check if it satisfies the equation.

Let's start by substituting 4-3i into each equation:

1) x^2 + 6x - 25 = 0
Plugging in 4-3i:
(4-3i)^2 + 6(4-3i) - 25 = 0
(16 - 24i + 9i^2) + (24 - 18i) - 25 = 0
(16 - 24i - 9) + (24 - 18i) - 25 = 0
(7 - 42i) - 1 - 25 = 0
7 - 42i - 1 - 25 = 0
-19 - 42i = 0

2) x^2 - 6x + 25 = 0
Plugging in 4-3i:
(4-3i)^2 - 6(4-3i) + 25 = 0
(16 - 24i + 9i^2) - (24 - 18i) + 25 = 0
(16 - 24i - 9) - (24 - 18i) + 25 = 0
(7 - 42i) - (1 - 18i) + 25 = 0
7 - 42i - 1 + 18i + 25 = 0
31 - 24i = 0

3) x^2 + 8x - 25 = 0
Plugging in 4-3i:
(4-3i)^2 + 8(4-3i) - 25 = 0
(16 - 24i +9i^2) + (32 - 24i) - 25 = 0
(16 - 24i - 9) + (32 - 24i) - 25 = 0
(7 - 42i) + (8 - 24i) - 25 = 0
7 - 42i + 8 - 24i - 25 = 0
-10 - 66i = 0

4) x^2 - 8x + 25 = 0
Plugging in 4-3i:
(4-3i)^2 - 8(4-3i) + 25 = 0
(16 - 24i +9i^2) - (32 - 24i) + 25 = 0
(16 - 24i - 9) - (32 - 24i) + 25 = 0
(7 - 42i) - (8 - 24i) + 25 = 0
7 - 42i - 8 + 24i + 25 = 0
24 - 18i = 0

After substituting the complex number 4-3i into each equation, we can see that only option 2) x^2 - 6x + 25 = 0 results in the equation being satisfied, since the expression simplifies to 31 - 24i which is equal to zero.

Therefore, the equation that has the complex number 4-3i as a root is option 2) x^2 - 6x + 25 = 0.

y the factors are (x-4+3i) and (x-4-3i) ???Can you explain please?

if 4-3i is a root, so is 4+3i

then factors are (x-4+3i) and (x-4-3i)

set (x-4+3i)(x-4-3i) = 0 and expand to see which of the equation you get.
(remember that i^2 = -1)