To determine which quadratic equation has roots of \(2+i\) and \(2-i\), we can use the fact that if a quadratic equation has roots \(r_1\) and \(r_2\), it can be expressed in the form:
\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]
First, we calculate the sum and product of the roots \(2+i\) and \(2-i\):
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Sum of the roots: \[ r_1 + r_2 = (2+i) + (2-i) = 4 \]
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Product of the roots: \[ r_1 \cdot r_2 = (2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5 \]
Now we can substitute these values into the quadratic equation format:
\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \] \[ x^2 - 4x + 5 = 0 \]
Now let's compare this with the given equations:
- Equation 1: \(x^2 - 4x + 5 = 0\) ✔️
- Equation 2: \(x^2 + 4x + 5 = 0\) ❌
- Equation 3: \(x^2 - 4x - 5 = 0\) ❌
The correct equation that has roots \(2+i\) and \(2-i\) is Equation 1: \(x^2 - 4x + 5 = 0\).