# Why is it that the roots of a quadratic is closest when the discriminant is smallest (but non-zero)?

My teacher told me earlier today that the roots of a quadratic are the closest when the discriminant is smallest, but non-zero. I wasn't able to understand why that is the case, however.

Is it correct to say that the discriminant is the distance between the roots? And since you can't have negative distance, that's why the roots are imaginary when the discriminant is negative?

• It is not correct to say that. However, if the discriminant $\Delta$ satisfies $0<\Delta <<1$, then the quadratic formula tells us that the two roots $m\pm n\sqrt{\Delta}$ are close, since $n\sqrt{\Delta}$ is very small. – Harry Oct 18 '17 at 16:26

If the roots coincide the discriminant is $0$ ($(x-r)^2=x^2-2rx+r^2$ has discriminant $0$) so it is natural to imagine that a very small discriminant means that the roots are close. However, the relation between the difference and the discriminant is not as simple as you might like.

If the quadratic is $ax^2+bx+c$ with discriminant $\Delta =b^2-4ac$ then the roots are $r_{\pm}=\frac {-b\pm \sqrt {\Delta}}{2a}$ so the difference is $$r_+-r_-=\frac {\sqrt {\Delta}}a$$

Thus $$\boxed {\Delta = a^2(r_+-r_-)^2}$$

if we have $ax^2+bx+c=0$, by completing the square we'll obtain $(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$. Taking square root from both sides, we get $\lvert x+\frac{b}{2a}\rvert=\frac{\sqrt{b^2-4ac}}{2\lvert a\rvert}$. From here we can see that two roots $x_1$ and $x_2$ will be equidistant from point $x=- \frac{b}{2a}$ and that distance is directly proportional to square root of discriminant. When discriminant is approaching 0, both roots approach $- \frac{b}{2a}$

This is a question with some interest and subtlety. Strictly the discriminant of a quadratic expression in one variable ought to be the square of the difference between the roots. e.g. for cubics and higher you take the product of the squares of differences of pairs of roots - for the three roots $a,b,c$ of a cubic it would be $(a-b)^2(b-c)^2(c-a)^2$.

However, the most usual use of discriminant for the quadratic $f(x)=ax^2+bx+c$ is $b^2-4ac$, which is the expression which appears under the square root sign in the quadratic formula. This is $a^2$ times the square of the difference between the roots.

Whichever way you compute it, if the discriminant is zero, the roots are equal. You have to adjust by the factor $a$ as appropriate to context to measure the difference (distance) between the roots using the square root of the discriminant.

Note: the fact that the discriminant is the square of the differences between the roots of a polynomial means that it is a symmetric function of the roots and can be expressed as a function of the coefficients.