how does (x-2-3i)(x-2+3i)=0 turns into (x-2)^2 - (3i)^2=0?
To start, let's multiply the given expression:
(x-2-3i)(x-2+3i)
Using the distributive property and the pattern (a-b)(a+b) = a^2 - b^2, we have:
= (x-2)^2 - (3i)^2
Now, let's simplify further:
= (x-2)^2 - 9i^2
Since i^2 is defined as -1, we can substitute -1 for i^2:
= (x-2)^2 - 9(-1)
Simplifying the expression, we have:
= (x-2)^2 + 9
Therefore, (x-2-3i)(x-2+3i) is equivalent to (x-2)^2 + 9 = 0.
To understand how the equation (x-2-3i)(x-2+3i) = 0 turns into (x-2)^2 - (3i)^2 = 0, we can simplify the left-hand side of the equation using the difference of squares formula.
1. Start with the equation (x-2-3i)(x-2+3i) = 0.
2. Multiply the terms using the foil method: (x-2)(x) + (x-2)(-3i) + (-2-3i)(x) + (-2-3i)(-3i) = 0.
3. Simplify each term: x^2 - 2x - 3ix - 2x + 4 + 6ix - 3ix + 6i^2= 0.
4. Combine like terms: x^2 - 4x + 4 + 6i^2 = 0.
5. Simplify i^2: since i^2 is defined as -1, the equation becomes x^2 - 4x + 4 - 6 = 0.
6. Simplify the equation: x^2 - 4x - 2 = 0.
Therefore, the equation (x-2-3i)(x-2+3i) = 0 simplifies to (x-2)^2 - (3i)^2 = 0 as x^2 - 4x - 2 = 0.