What is the role played by the difference of the two roots in solving a quadratic equation?

Question

I have to write an essay and this is the main question of the essay, and I'm massively stuck.

All I have so far is the difference of squares theorem, which I think might just be massively irrelevant. The module is Using Radicals to Solve and I not only don't understand the question, I can't find any relevant resources.

I just don't think I'm going in the right direction so it would be so so so appreciated if somebody could give me their interpretation of what I'm being asked, or even point me in the right direction of some online resources etc.

Thank you so much

Answer 1

The roots are given by the average of the roots (which by symmetry is the abscissa of the extremum) plus or minus the half difference.

Answer 2

You might find the video at the bottom of my post useful; he explains in good detail how the difference in two roots can be found by using the following equation:

$$m\pm\sqrt{m^2-p}$$

$m$: The midpoint between the roots.

$p$: The constant which has been adjusted.

$\sqrt{m^2-p}$ is the distance from the midpoint to the nearest root; obviously there will be a $\pm$ before it as since there are two roots, one will occur in the negative $x$ direction and one in the positive $x$. The video below will explain in more detail; go to the 16th minute if you need an answer quick.

Quadratic Equation Video

Answer 3

If f$x_1,x_2$ are the two roots of a quadratic polynomial $$X^2+pX+q,$$ then their difference is $$\delta:=x_1-x_2,$$ of course. The essay could end here, but note that the roots do not form an ordered pair, i.e., we might as well swap their order, which produces a different $\delta$, namely $x_2-x_1$. This is almost the same, namely minus the original $\delta$. What do we do to get rid of the uncertainty of the sign? We square! So the more natural expression in terms of difference of roots is $D:=\delta^2=(x_1-x_2)^2$. This $D$ does not depend on the order of $x_1,x_2$; we say that it is symmetric in $x_1,x_2$. As it expands as $D=x_1^2+x_2^2-2x_1x_2$, this is clear anyway.

Fun fact: Every symmetric polynomial in two variables $x_1,x_2$ can be expressed as a polynomial in the elementary symmetric polynomials $x_1+x_2$ and $x_1x_2$ (this generalizes to more than two variables with correspondingly more elementary symmetric polynomials). Concretely for $D$, this is not hard to do. In fact, in order to handle the $x_1^2$ occurring in $D$, we should use $(x_1+x_2)^2$. This leaves us with $D-(x_1+x_2)^2=-4x_1x_2$ and we are done.

Also note that $$ (X-x_1)(X-x_2)=X^2-(x_1+x_2)X+x_1x_2$$ and by comparing coefficients, we find $x_1+x_2=-p$ and $x_1x_2=q$. So now now our investigations of the difference of roots led us directly to the discovery that $$ (x_1-x_2)^2=D=(x_1+x_2)^2-4x_1x_2=p^2-4q$$ and so $$\delta=\pm\sqrt{p^4-4q}. $$ Knowing $a_1+x_2=p$ and $x_1-x_2=\delta$, it is a trivial task to find $x_1$ and $x_2$ themselves.

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For a cubic polynomial, we would investigate $(x_1-x_2)^2(x_2-x_3)^2(x_1-x_3)^2$ instead, which is again symmetric and very helpful.

Answer 4

You can't find the roots themselves from just the difference of the roots.

With the sum of the roots and their product, you can find the roots (such as by Vieta's formulas), but the difference of the roots and their product will work just fine.

Specifically, if $\alpha$ and $\beta$ are the two roots, then with:

$$\alpha - \beta = p \tag{1}$$ $$\alpha \beta = q \tag{2}$$

set $\beta = -\gamma$ and you have $\alpha + \gamma = p$, $\alpha \gamma = -q$ which is basically Vieta's formulas all over again. This gives you $b$ and $c$ in terms of $a$ in the quadratic equation $ax^2+bx+c = 0$ which you can then solve.