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

I have no clue how to do it!!!PLEASE HELP!!!Please guide by step by step!!THANKS A LOT~!!!!

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To find the equation with the complex number 4-3i as a root, we can use the fact that complex roots of a quadratic equation come in conjugate pairs.

Step 1: Use the root to find the corresponding binomial factors.
A complex root in the form a+bi, where a and b are real numbers, will have a conjugate of a-bi as the other root. In this case, the conjugate of 4-3i is 4+3i.

Step 2: Write the binomial factors.
For the complex root 4-3i, the associated binomial factors are (x - (4 - 3i)) and (x - (4 + 3i)).

Step 3: Multiply the binomial factors.
(x - (4 - 3i))(x - (4 + 3i))

Step 4: Simplify the expression.
Using the difference of squares formula, we can expand and simplify the expression:
(x - 4 + 3i)(x - 4 - 3i) = (x - 4)^2 - (3i)^2 = x^2 - 8x + 16 - 9 = x^2 - 8x + 7

So, the equation that has the complex number 4-3i as a root is option (4) x^2 - 8x + 25 = 0.