# 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. Oct 7, 2020 at 1:55
• This is for equations whose coefficients are real. Oct 7, 2020 at 1:56
• According to the quadratic formula, the roots are complex when $b^2-4ac<0$ Oct 7, 2020 at 1:56

This only applies when all the coefficients are real numbers.

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? Oct 7, 2020 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. Oct 7, 2020 at 2:08

To pursue your idea of using a "difference of two squares", we can start by writing the quadratic equation in "vertex form" thus, $$ax^2 \ + \ bx \ + \ c \ \ = \ \ a·\left(x + \frac{b}{2a}\right)^2 \ + \ \left(\underbrace{c \ - \ \frac{b^2}{4a}}_{k} \right)$$ $$= \ \ a·\left(x + \frac{b}{2a}\right)^2 \ - \ \left( \frac{b^2}{4a} \ - \ c \right) \ \ = \ \ 0 \ \ ,$$ with the polynomial coefficients all being real numbers.

The equation will have non-real roots if the vertex is "above" the $$\ x-$$axis $$\ ( k > 0 ) \$$ for $$\ a > 0 \ \$$ and "below" that axis $$\ ( k < 0 ) \$$ when $$\ a < 0 \ \ ;$$ this is equivalent to the statement $$\ b^2 - 4ac < 0 \$$ in regard to the discriminant of the quadratic polynomial.

The difference of terms can then be factored as $$a \ · \ \left[ \ \left(x + \frac{b}{2a}\right)^2 \ - \ \left( \ \sqrt{ \underbrace{\frac{b^2 \ - \ 4ac}{4a^2}}_{< \ 0}} \ \right)^2 \ \right]$$ $$= \ \ a \ · \left[ \ \left(x + \frac{b}{2a}\right) \ + \ i· \frac{\sqrt{4ac \ - \ b}} {2a} \ \right] · \left[ \ \left(x + \frac{b}{2a}\right) \ - \ i· \frac{\sqrt{4ac \ - \ b}} {2a} \ \right] \ \ = \ \ 0 \ \ ,$$ from which it follows that the zeroes of the quadratic polynomial form the "complex conjugate pair" $$\ -\frac{b}{2a} \ \pm \ i· \frac{\sqrt{4ac \ - \ b}} {2a} \ \ .$$