Discriminant of a Quadratic Polynomial

Question

I am a little confused about how the author came to this discriminant of this quadratic polynomial. I understand that the discriminant comes from solving for a variable, and the discriminant is whatever is under the square root. Below is information directly from the PDF:

(1) $f(y) = (y^2 − v^2)(y − u)$,

with rational u, v. Indeed, to get all the roots of f rational, it is enough to assume that u, v are rational.

The quadratic polynomial:

(2) $f'(y) = 3y^2 − 2yu − v^2$

has rational roots, if and only if its discriminant $4(u^2 + 3v^2)$ is the square of a rational number.

How is the discriminate $4(u^2 + 3v^2)$? When I solve for $y$ in equation (2), I get $(u^2 + 3v^2)$ under the square root. I am guessing that the author is assuming that $4(u^2 + 3v^2)$ is the discriminant?

Answer 1

We have $f'(y)=3y^2-2uy-v^2$. it is a quadratic equation in $y$. so discriminant is $$\Delta=(-2u)^2-4(3)(-v^2)=4u^2+4(3v^2)=4(u^2+3v^2)$$

Answer 2

There are many ways to find the roots of a quadratic polynomial. First thing you want to figure out is how many real roots it has. How to do that? The discriminant tells you this exact type of information.

First, I would like to figure out the roots of the polynomial $f(x) = ax^2 + bx + c$. How can we figure out the roots? We can easily do that by setting $f(x) = 0$. Thus: $$\begin{align*} f(x) = 0\\ a x^2 + b x + c = 0 \end{align*}$$

Solving for $x$ in there can be done in many ways, but I will go with the completing the square idea. First step is to divide the entire equation by $a$ and then move constant terms to the right hand side: $$\begin{align*} a x^2 + b x + c & = 0\\ x^2 + \frac{b}{a} x + \frac{c}{a} & = 0\\ x^2 + \frac{b}{a} x & = -\frac{c}{a} \end{align*}$$

Now let us think about $(u + v)^2$. If you expand it, you will get: $$\begin{align*} (u + v)^2 & = u^2 + uv + uv + v^2\\ (\color{green}u + \color{blue}v)^2 & = \color{green}u^2 + 2\color{green}u\color{blue}v + \color{blue}v^2 \end{align*}$$

I will now highlight a few terms in the equation for the roots and also do a small manipulation. Maybe you will few comfortable about noticing a few things. $$\begin{align*} \color{green}x^2 + \color{blue}{\frac{b}{a}}\color{green}x & = -\frac{c}{a}\\ \color{green}x^2 + \color{red}2\color{green}x\color{blue}{\frac{b}{\color{red}{2}a}} & = -\frac{c}{a}\\ (\color{green}x)^2 + 2(\color{green}x)\left(\color{blue}{\frac{b}{2a}}\right) & = -\frac{c}{a} \end{align*}$$

If you look at the equation we got above, you will notice it is almost a perfect square. However, it is missing a term and we can add it to both sides. $$\begin{align*} (\color{green}x)^2 + 2(\color{green}x)\left(\color{blue}{\frac{b}{2a}}\right) \color{red}{+ \left(\frac{b}{2a}\right)^2} & = -\frac{c}{a} \color{red}{+ \left(\frac{b}{2a}\right)^2}\\ (\color{green}x)^2 + 2(\color{green}x)\left(\color{blue}{\frac{b}{2a}}\right) + \left(\color{blue}{\frac{b}{2a}}\right)^2 & = -\frac{c}{a} + \left(\color{blue}{\frac{b}{2a}}\right)^2\\ \left(\color{green}x + \color{blue}{\frac{b}{2a}}\right)^2 & = \left(\color{blue}{\frac{b}{2a}}\right)^2 - \frac{c}{a} \end{align*}$$

We can simplify the above but first we must notice that the solutions to $x^2 = n^2$ can be both $x = -n$ and $x = n$. Thus: $$\begin{align*} x^2 = n^2\\ x = \pm \sqrt{n^2} \end{align*}$$

Going back to our problem: $$\begin{align*} \left(x + \frac{b}{2a}\right)^2 & = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}\\ \left(x + \frac{b}{2a}\right)^2 & = \frac{b^2}{4a^2} - \frac{c}{a}\\ \left(x + \frac{b}{2a}\right)^2 & = \frac{b^2}{4a^2} - \frac{\color{red}{4a}c}{\color{red}{4a}a}\\ \left(x + \frac{b}{2a}\right)^2 & = \frac{b^2}{4a^2} - \frac{4ac}{4a^2}\\ \left(x + \frac{b}{2a}\right)^2 & = \frac{b^2 - 4ac}{4a^2}\\ x + \frac{b}{2a} & = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}\\ x + \frac{b}{2a} & = \pm \frac{\sqrt{b^2 - 4ac}}{2a}\\ x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*}$$

We have successfully found what gives us the roots of a quadratic polynomial. However, there is something which tells us the nature of these roots: the discriminant ($\Delta$).

In the formula we derived, the conditions for the roots to be real or not, is bounded to the square root. Thus, we define the inside of the square root as the discriminant of the polynomial. $$\boxed{\Delta = b^2 - 4ac}$$

In the formula you provided, we have: $$\begin{align*} f(y) & = 3y^2 - 2uy - v^2\\ f(y) & = \color{red}{3}y^2 \color{red}{-2u}y \color{red}{-v^2}\\ a & = 3\\ b & = -2u\\ c & = -v^2 \end{align*}$$

Therefore, your discriminant $\Delta$ will be: $$\begin{align*} \Delta & = (-2u)^2 - 4(3)(-v^2)\\ \Delta & = 4u^2 + 4(3v^2)\\ \Delta & = 4(u^2 + 3v^2) \end{align*}$$

Which is what you originally asked.