Started complex today, I think this works an alternative though
ax^2 + bx+ c. Since one of the roots are complex, the other must complex also.
Then the complex root alpha = (-b +- i(b^2 - 4ac)^(1/2))/2
Obviously then the other root has the sign switched in front of the 'i', which is its conjugate.
The question in itself is to prove the conjugate root theorem for quadratics. So that bolded part cannot be claimed just yet.
Your proof is valid in its current form, but it is a very weak proof. It seems like your approach is to use the quadratic formula to explicitly find the roots and then to show that the roots are conjugates of each other.
Now, if the coefficients are real, then this result is obviously true.
However, it has the following weaknesses.
- This method only works for quadratics, since you know an explicit formula for solving them. What if the polynomial was a cubic instead?
- This method doesn't demonstrate clearly why it is important to have the coefficients being real.
- Proving the converse (excluding real polynomials multiplied by a non-real constant) is difficult for higher order polynomials.
=a\alpha^2 + b\alpha + c=0 \\\\ $In order to prove that the conjugate $ \bar{\alpha} $ is also a root, we must prove that $ P(\bar{\alpha}}) = 0 \\\\ P(\bar{\alpha}}) = a\bar{\alpha}^2 + b \bar{\alpha}+ c = \bar{a} \bar{\alpha}}^2 + \bar{b} \bar{\alpha} + \bar{c} = \bar{a\alpha^2} + \bar{b\alpha} + \bar{c} = \bar{a\alpha^2 + b\alpha + c} = \bar{0} = 0 \\\\ $ Note that we have used the following three properties:$ \\\\ $1. If $ k $ is a real number, then $ k=\bar{k} \\\\ $2. $ \bar{z_1}\bar{z_2} = \bar{z_1 z_2} \\\\ $3. $ \bar{z_1} + \bar{z_2} = \bar{z_1+z_2})
=a\alpha^2 + b\alpha + c=0 \\\\ $In order to prove that the conjugate $ \bar{\alpha} $ is also a root, we must prove that $ P(\bar{\alpha}}) = 0 \\\\ P(\bar{\alpha}}) = a\bar{\alpha}^2 + b \bar{\alpha}+ c = \bar{a} \bar{\alpha}}^2 + \bar{b} \bar{\alpha} + \bar{c} = \bar{a\alpha^2} + \bar{b\alpha} + \bar{c} = \overline{a\alpha^2 + b\alpha + c} = \bar{0} = 0 \\\\ $ Note that we have used the following three properties:$ \\\\ $1. If $ k $ is a real number, then $ k=\bar{k} \\\\ $2. $ \bar{z_1}\bar{z_2} = \overline{z_1 z_2} \\\\ $3. $ \bar{z_1} + \bar{z_2} = \overline{z_1+z_2})
