Question
explain why complex solutions come in conjugate pairs. Consider the structure of the quadratic formula in your response.(4 points)
Answers
GPT-4o mini
Complex solutions come in conjugate pairs due to the properties of polynomial equations and the nature of complex numbers. To explain this, we can look at the quadratic formula and how it derives complex solutions.
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a\), \(b\), and \(c\) are coefficients of the quadratic equation \(ax^2 + bx + c = 0\).
1. **Discriminant Analysis**: The expression under the square root, \(b^2 - 4ac\), is known as the discriminant. If \(b^2 - 4ac < 0\), the square root of a negative number emerges when calculating the solutions, resulting in complex numbers. Specifically, we have:
\[
\sqrt{b^2 - 4ac} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2}
\]
where \(i\) is the imaginary unit.
2. **Structure of the Solutions**: When we apply this to the quadratic formula, for a negative discriminant, we end up with:
\[
x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}
\]
The solutions can be expressed as:
\[
x_1 = \frac{-b}{2a} + i\frac{\sqrt{4ac - b^2}}{2a}
\]
\[
x_2 = \frac{-b}{2a} - i\frac{\sqrt{4ac - b^2}}{2a}
\]
3. **Conjugate Pair Explanation**: The two solutions, \(x_1\) and \(x_2\), are now in the form of a complex number and its conjugate. The conjugate is formed by changing the sign of the imaginary part. This is a fundamental property of complex numbers: if a polynomial with real coefficients has a complex solution, its conjugate must also be a solution. This is because when the polynomial is factored or expanded, any non-real roots must occur in pairs to keep the coefficients real.
4. **Conclusion**: Therefore, when we encounter complex roots as solutions to a quadratic equation (or any polynomial with real coefficients), they always arise as conjugate pairs due to the necessity of maintaining the realness of the coefficients in the polynomial. This symmetry ensures that complex solutions inherently exist in pairs, preserving the algebraic structure of the equation.
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a\), \(b\), and \(c\) are coefficients of the quadratic equation \(ax^2 + bx + c = 0\).
1. **Discriminant Analysis**: The expression under the square root, \(b^2 - 4ac\), is known as the discriminant. If \(b^2 - 4ac < 0\), the square root of a negative number emerges when calculating the solutions, resulting in complex numbers. Specifically, we have:
\[
\sqrt{b^2 - 4ac} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2}
\]
where \(i\) is the imaginary unit.
2. **Structure of the Solutions**: When we apply this to the quadratic formula, for a negative discriminant, we end up with:
\[
x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}
\]
The solutions can be expressed as:
\[
x_1 = \frac{-b}{2a} + i\frac{\sqrt{4ac - b^2}}{2a}
\]
\[
x_2 = \frac{-b}{2a} - i\frac{\sqrt{4ac - b^2}}{2a}
\]
3. **Conjugate Pair Explanation**: The two solutions, \(x_1\) and \(x_2\), are now in the form of a complex number and its conjugate. The conjugate is formed by changing the sign of the imaginary part. This is a fundamental property of complex numbers: if a polynomial with real coefficients has a complex solution, its conjugate must also be a solution. This is because when the polynomial is factored or expanded, any non-real roots must occur in pairs to keep the coefficients real.
4. **Conclusion**: Therefore, when we encounter complex roots as solutions to a quadratic equation (or any polynomial with real coefficients), they always arise as conjugate pairs due to the necessity of maintaining the realness of the coefficients in the polynomial. This symmetry ensures that complex solutions inherently exist in pairs, preserving the algebraic structure of the equation.