# Using quadratic formula to prove that if quadratic equation has any non-real roots, then there must be 2 roots which are conjugates of each other?

How to use quadratic formula to prove that if a quadratic equation has any non-real roots, then there must be 2 roots which are conjugates of each other?

I figure this might have something to do with difference of two squares?

$$(a-bi)(a+bi)$$ as it would mean the roots are conjugates of each other

Or maybe the discriminant formula $$b^2-4ac > 0$$? But I do not know how to formally write this proof.

Any help would be much appreciated!

• You need an extra condition: the coefficients of the polynomial need to be real. Otherwise we could write for example $(z - i)(z - 2i) = z^2 - 3i - 2$ for a polynomial with two non-real roots which are not complex conjugates. – Joppy Oct 7 '20 at 1:55
• This is for equations whose coefficients are real. – Robert Israel Oct 7 '20 at 1:56
• According to the quadratic formula, the roots are complex when $b^2-4ac<0$ – J. W. Tanner Oct 7 '20 at 1:56

For $$ax^2 + bx + c =0$$ you have $$x = \frac{-b\pm \sqrt{b^2-4ac} } {2a}$$.
For one root to be non-real you need the discriminant $$b^2-4ac<0$$. If you let $$D^2 =4ac-b^2$$, then the roots can now be written as $$x = - \frac{b}{2a}\pm i\frac{D}{2a}$$, which are, by definition (same real part, non-zero imaginary part of opposite sign), conjugate complex numbers.
• I don't understand, where do we get the $D^2$ from? Is that the discriminant? – user780357 Oct 7 '20 at 2:03
• $D$ is a new variable I introduced that represents the square root of the negative of the discriminant. The negative sign of the discriminant needed to be reversed before a (real) square root could be taken. This is just a neater presentation. – Deepak Oct 7 '20 at 2:08