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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!

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    $\begingroup$ 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. $\endgroup$
    – Joppy
    Commented Oct 7, 2020 at 1:55
  • $\begingroup$ This is for equations whose coefficients are real. $\endgroup$
    – Robert Israel
    Commented Oct 7, 2020 at 1:56
  • $\begingroup$ According to the quadratic formula, the roots are complex when $b^2-4ac<0$ $\endgroup$
    – J. W. Tanner
    Commented Oct 7, 2020 at 1:56

2 Answers 2

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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.

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  • $\begingroup$ I don't understand, where do we get the $D^2$ from? Is that the discriminant? $\endgroup$
    – user780357
    Commented Oct 7, 2020 at 2:03
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    $\begingroup$ $D$ 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. $\endgroup$
    – Deepak
    Commented Oct 7, 2020 at 2:08
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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} \ \ . $

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