# Discriminant of General Quadratic Equation

1. Jul 13, 2009

### DJ24

I understand how the discriminant, $$b^{2}-4ac$$, comes from in the quadratic equation $$ax^{2}+bx+c=0$$, but how does it come from the general quadratic equation $$ax^{2}+bxy+cy^{2}+dx+ey+f=0$$ ?

2. Jul 14, 2009

### HallsofIvy

Staff Emeritus
3. Jul 14, 2009

### DJ24

I still do not see the connection between $$b^{2}-4ac$$ and the classification of a conic.

4. Jul 14, 2009

### symbolipoint

The discriminant expression occurs in the completion of the square and solution process for a quadratic equation. One of the requirements is that one accepts the connection between conic sections and the use of the distance formula definitions for parabola, ellipse, circle, and hyperbola.

5. Jul 14, 2009

### DJ24

I know where the discriminant comes from in the quadratic formula of which involves only x, but I don't see how it comes from the irreducible general quadratic equation of which involves x and y.

6. Jul 14, 2009

### symbolipoint

A student or other interested person could take the equation for an unrotated conic section for which the xy term would be zero, and solve for x in terms of y. A discriminant expression would occur; actually, I should try this just to be sure, although I feel that by doing so I would find what I expect.

7. Jul 14, 2009

### HallsofIvy

Staff Emeritus
Yes, that, together with a number of different ways of looking at it, was explained to DJ24 the first time he posted this question. Since he says he did not understand them, I see no reason to think he will understand this time around.

8. Jul 14, 2009

### DJ24

I narrowed down my previous question and reposted it because I felt like the last one died. Also, my previous question was not a request of an explanation of the determinant's derivation, but rather the connection between it and conic classification.

I feel like there may be a simple proof of how b2-4ac can be derived from the general quadratic equation just as it is derived from the quadratic equation with only x, and was hoping someone could show it to me. I am not satisfied until I, if possible, have an intuitive and thorough understanding of the matter; I do not just accept plain facts of which I do not understand.

9. Jul 14, 2009

### symbolipoint

Are you asking for a geometric interpretation for the discriminant? The symbolism occurs in the derivation for solution of x or y. This then may have meaning regarding restrictions on Real values for x or y, since b2 - 4ac being negative indicates an imaginary number. I am about at my limit for what I can tell you about this (and the b2-4ac is for solution of x for a parabola; I have not yet tried to perform the derivation on the general quadratic equation for x and for y). This would be for someone else to explain.

10. Jul 14, 2009

### DJ24

I am looking for more of an algebraic proof.

11. Jul 14, 2009

### trambolin

What kind of algebraic proof are you looking for?

12. Jul 14, 2009

### DJ24

Is the eigenvalue/vector explanation the only one, though? Isn't there a simple algebraic rearrangement of the general quadratic equation that results in displaying the discriminant?-such as in the quadratic formula of which only involves the variable x?

13. Jul 14, 2009

### trambolin

You mean something like this?

\begin{align*} ax^2+bx+c &= 0, a\neq 0\\ x^2+\frac{b}{a}x+\frac{c}{a} &= 0\\ x^2+\frac{b}{a}x+ (\frac{b}{2a})^2 &= (\frac{b}{2a})^2-\frac{c}{a}\\ \left( x+ \frac{b}{2a} \right) &= \pm\frac{\sqrt{b^2-4ac}}{2a}\\ x &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align*}

Edit : This trick is completion of squares as suggested before. You can complete the square for the general one which is not so fun to type in latex if that is what you mean...Please be specific about what you ask for. Saying "no that is not what I want" does not lead to fruitful posts.

Addition to HallsofIvy's representation the full general quadratic form can also be written as
$$\begin{bmatrix} x & y & 1\end{bmatrix}\begin{bmatrix} A & \frac{B}{2} &\frac{D}{2}\\ \frac{B}{2} & C &\frac{E}{2}\\ \frac{D}{2} &\frac{E}{2} &F\end{bmatrix}\begin{bmatrix} x \\ y\\ 1\end{bmatrix} = 0$$

You can go for the determinant of this... If there are no eigenvalues at zero, can this be possible for some (x,y)?

Last edited: Jul 15, 2009
14. Jul 15, 2009

### HallsofIvy

Staff Emeritus
Start with $Ax^2+ Bxy+ Cy^2$ and complete the square just like with the single variable problem:
$$A(x^2+ (B/A)xy+ (By/2A)^2- (By/2A)^2+ Cy^2= A(x- By/2A)^2+ y^2(C- (B/2A)^2$$
Assuming A is positive (which you could guarentee by multiplying the entire equation by -1 if necessary), that will be a "sum of squares", and so the graph is an ellipse or circle, if and only if $C- B^2/4A^2> 0$ or $B^2- 4AC< 0$.

It will be a "difference of squares", and so the graph is a hyperbola, if and only if $B^2- 4AC> 0$ and will have only one square, and so the graph is a parabola if and only $B^2- 4AC= 0$.

15. Jul 16, 2009

### DJ24

I see how the discriminant arises by completing the square: $$(2ax+by)^{2}=(b^{2}-4ac)y^{2}$$

But how does it relate to a parabola when $$b^{2}-4ac=0$$?