Discriminant of a quadratic equation

Question

I am trying to write a mathematical paper and I am including the following paragraph about the discriminant of the quadratic equation. I am trying to wrap my head around this concept and I wanted to make sure my logic was correct.

The discriminant reveals the nature of the roots of a quadratic equation given that $a, b$ and $c$ are rational numbers. When calculated it determines the number of real roots, or in other words, the number of x-intercepts, associated with a quadratic equation. For example, if $b^2 - 4ac = 0$, then the quadratic equation will have exactly one real number solution (with duplicity.) In other words, there is one solution that is repeated resulting in the curve intersecting the $x$-axis at one point. If $b^2 - 4ac > 0$ (in other words the discriminant is a positive number), then the quadratic equation will have exactly two real number solutions resulting in the curve intersecting the $x$-axis at two distinct points. If $b^2 - 4ac < 0$ (in other words the discriminant is a negative number), then the quadratic equation will have no real solutions. Instead there will be two complex number solutions resulting in the curve having no intersection points on the $x$-axis.

Answer 1

If a,b,c will be real numbers, discriminant will just tell whether the roots will be real or complex and whether they will be equal or unequal. If a, b, c will be rational numbers discriminant will also tell us about rationality or irrationality of roots.

Answer 2

You are correct in your logic.

• If the discriminant is equal to $0$ then the equation has only $1$ real root, because the equation touches the $x$-axis at that one point.
• If the discriminant is greater than $0$ then the equation has $2$ real roots, because the equation cuts the $x$-axis at $2$ points.
• If the discriminant is less than $0$ then the equation has no real roots, because the equation neither touches the $x$-axis or goes below it.

Answer 3

Descartes’ big idea was indeed to add $y$ to an equation in order to see what happens; so from $ax^2+bx+c=0$ he considered instead $$ y=ax^2+bx+c $$ which represents a parabola (he could prove this). We can rewrite it as $$ y=a\left(x^2+2\frac{b}{2a}x+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\right) $$ (we can see here the completion of the square). This becomes $$ \frac{y}{a}=\left(x+\frac{b}{2a}\right)^{\!2}-\frac{b^2-4ac}{4a^2} $$ Setting $X=x+\frac{b}{2a}$ doesn't change the number of solutions, because it just shifts them and the curve is still a parabola: $$ \frac{y}{a}=X^2-\frac{b^2-4ac}{4a^2} $$

Under the non restrictive assumption that $a>0$, we now clearly see the three possible cases:

1. if $b^2-4ac<0$, we have $y>0$ for every $X$ (so for every $x$);
2. if $b^2-4ac=0$, we have $y=0$ only for $X=0$ (the parabola touches the $X$-axis);
3. if $b^2-4ac>0$, the parabola intercepts twice the $X$-axis (and the $x$-axis as well), because $y<0$ for $X=0$, but $y>0$ for “large positive or negative $X$”.

Analyzing higher degree equations and also irrational ones with such geometric methods, could (and indeed does) give information about the solutions. Descartes’ folium was studied in connection with degree three equations. At the end it turned out that Descartes’ method is better in applying algebra to the study of geometry, rather than the converse, but this doesn't diminish at all his big contribution to mathematics.

The case of quadratic equations is easy from the purely algebraic point of view, because the equation becomes, with the same steps, $$ X^2=\frac{b^2-4ac}{4a^2} $$ but having a geometric insight helps; for instance, when the coefficients $b$ and $c$ depend on a parameter.