Math Help - A fundamental question

1. A fundamental question

I have a very fundamental question regarding quadratic equations/inequations. I know that the discriminant determines whether the roots are real or imaginary, but my question is beyond that :

What is the role of discriminant in a quadratic equation/inequation ?

1) Does it provide the Range of the solution set of the quadratic equation/inequation ?

2) Does it provide the Range of values attained by the quadratic equation/inequation ?

Is it both the above ?

2. Re: A fundamental question

If you are talking about ax^2+bx+c > 0, there are 2 possibilities.
If delta < 0, ax^2+bx+c=0 has no real root, then yes, the negative discriminant tells you that ax^2+bx+c > 0 for all real numbers x.
If delta = 0, there is one real root, and the inequality stands for anything but that number. To find the specific set, though, you have to actually solve ax^2+bx+c=0.

3. Re: A fundamental question

In this context I will post a question which will bring the concept to light:

If x,y,z are real numbers such that x+y+z=4 and x2+y2+z2=6, then x,y,z lie in ?

( I guess this question is asking for the Range of the solution set x,y and z or something else ? Or, maybe I am missing something.)

4. Re: A fundamental question

Is the above question about the Range of the solution set ?

5. Re: A fundamental question

Originally Posted by SheekhKebab
I have a very fundamental question regarding quadratic equations/inequations. I know that the discriminant determines whether the roots are real or imaginary, but my question is beyond that :

What is the role of discriminant in a quadratic equation/inequation ?

1) Does it provide the Range of the solution set of the quadratic equation/inequation ?

2) Does it provide the Range of values attained by the quadratic equation/inequation ?

Is it both the above ?
I would answer the question this way.

The discriminant ALONE does NOT determine the range of the real function $ax^2 + bx + c.$

The discriminant alone does determine the nature of the zeroes of that real function: whether the zeroes are real or complex and whether the real roots (if there are any) are distinct.

What determines the range of that real function is the sign of a and the value of $-\ \dfrac{b^2 - 4ac}{4a}.$

So the discriminant is a component of what determines the range of $ax^2 + bx + c,$ not the sole determinant.

You may wonder where $-\ \dfrac{b^2 - 4ac}{4a}$ comes from.

The local extremum of the real function $ax^2 + bx + c$ occurs if $x = -\ \dfrac{b}{2a}.$

So the value of the function at that extremum is

$a\left(-\ \dfrac{b}{2a}\right)^2 + b\left(-\ \dfrac{b}{2a}\right) + c = \dfrac{b^2}{4a} - \dfrac{b^2}{2a} + c = \dfrac{b^2}{4a} - \dfrac{2b^2}{4a} + \dfrac{4ac}{4a} = -\ \dfrac{b^2 - 4ac}{4a}.$

6. Re: A fundamental question

Originally Posted by SheekhKebab
In this context I will post a question which will bring the concept to light:

If x,y,z are real numbers such that x+y+z=4 and x2+y2+z2=6, then x,y,z lie in ?

( I guess this question is asking for the Range of the solution set x,y and z or something else ? Or, maybe I am missing something.)
I don't think the concepts of range and domain are directly pertinent to this problem, which is about equations, not functions. I think, however, that there is a somewhat similar concept involved, namely the bounds on the set of real solutions.

I do not see what the concept of the discriminant has to do with this problem at all.

It is trivially easy to show that $-\ \sqrt{6} \le x \le \sqrt{6}\ and\ -\ \sqrt{6} \le y \le \sqrt{6}\ and\ -\ \sqrt{6} \le z \le \sqrt{6}.$

7. Re: A fundamental question

An interesting solution:-

x+y+z=4 and x2+y2+z2=6,

y+z=4-x ...............(i)

and y2+z2=6-x2

yz=$\frac{1}{2}$[(y+z)2-(y2+z2)]

yz=x2-4x+5,.................(ii)

Hence, y and z are roots of:

t2-(4-x)t + (x2-4x+5)=0

Since the roots y and z are real, hence D>=0:

i.e (4-x)2-4(x2-4x+5)>=0

or (3x-2)(x-2)<=0

x = [ $\frac{2}{3}$,2]

and by symmetry y and z are also [$\frac{2}{3}$,2]

Now, you can clearly see that the Discriminant indeed gives the range of solution set x,y,z. Or am I missing something ?

8. Re: A fundamental question

Another question of this type-

k=$\frac{x^2-x+1}{x^2+x+1}$ [x belongs to real number]

kx2+kx+k=x2-x+1

(k-1)x2+(k+1)x+k-1=0

Since,x is real D>=0,

i.e (k+1)2-4(k-1)2>=0

on solving,we get: $\frac{1}{3}$<=k<=3, which is the range of the given expression.

So you can clearly see that the discriminant provides us the range of the solution set as well as the range of the expression. Please inform if I am missing something or if there is something wrong with my conception .

9. Re: A fundamental question

One thing to be noticed in the above expressions is that non of them is in the standard format of quadratic expression i.e
ax2+bx+c. One is a quadratic in 3 variables and other is a rational quadratic expression. So can we make a generalization that the Discriminant can give us the range of the solution set and the expression when expression is a non-standard quadratic expression? And which has been converted into a quadratic expression.
Also, that,in case of a standard quadratic expression the Discriminant alone does not gives us the range ?

10. Re: A fundamental question

Originally Posted by SheekhKebab
An interesting solution:-

x+y+z=4 and x2+y2+z2=6,

y+z=4-x ...............(i)

and y2+z2=6-x2

yz=$\frac{1}{2}$[(y+z)2-(y2+z2)]

yz=x2-4x+5,.................(ii)

Hence, y and z are roots of:

t2-(4-x)t + (x2-4x+5)=0

Since the roots y and z are real, hence D>=0:

i.e (4-x)2-4(x2-4x+5)>=0

or (3x-2)(x-2)<=0

x = [ $\frac{2}{3}$,2]

and by symmetry y and z are also [$\frac{2}{3}$,2]

Now, you can clearly see that the Discriminant indeed gives the range of solution set x,y,z. Or am I missing something ?
As far as I can see, your argument is sound but very confusingly presented. For example, you never define the variable t, and merely assert that
t^2 + t(y + z) + yz = 0.

Let's try it this way.

$A:\ (-y)^2 + (-y)(y + z) + yz = y^2 - y^2 - yz + yz = 0.$ So far so good; this is true for any numbers y and z.

$B:\ (y + z)^2 - 2yz = y^2 + 2yz + z^2 - 2yz = y^2 + z^2.$ Again true for any numbers y and z.

$C:\ x + y + z = 4 \implies (y + z) = (4 - x).$ Even I can understand that.

$x^2 + y^2 + z^2 = 6 \implies y^2 + z^2 = 6 - x^2 \implies (y + z)^2 - 2yz = 6 - x^2 \implies (4 - x)^2 - 2yz = 6 - x^2 \implies$

$D:\ yz = \dfrac{16 - 8x + x^2 - 6 + x^2}{2} = x^2 - 4x + 5.$

Putting all that together we get

$(-y)^2 + (-y)(y + z) + yz = 0 \implies y^2 - y(4 - x) + x^2 - 4x + 5 = 0.$

As you say, y is real so $(x - 4)^2 - 4 * 1 * (x^2 - 4x + 5) \ge 0.$

As everyone knows, the discriminant of a quadratic does determine whether the solution to the quadratic is real. What I find confusing is the following. The word "range" is ambiguous because the range of a real solution set and the range of a real function refer to different things. And the discriminant that is relevant is not necessarily the discriminant of any expression in the original equations. So "the" discriminant is not specified; it must be constructed case by case.

What I think you are saying is this:

Given a system of (n - 1) simultaneous independent and consistent polynomial equations involving n variables, where every polynomial is of degree less than 3 and at least one polynomial is of degree 2, there exists at least one quadratic the discriminant of which contains only one variable and which therefore bounds the possible real values of that variable, which in turn bounds the possible real values of all the other variables.

That sounds plausible to me, but I would hate to try to prove it.

11. Re: A fundamental question

Thanks JeffM, but what about the rational expression in the other question I have posted. You can see that the Discriminant does gives the range of the expression k.

12. Re: A fundamental question

Originally Posted by SheekhKebab
Thanks JeffM, but what about the rational expression in the other question I have posted. You can see that the Discriminant does gives the range of the expression k.
I am sorry to be argumentative, but you have never explained what proposition or propositions you are asking about. You keep talking about "THE" discriminant.

You give as an example $k = \dfrac{x^2 - x + 1}{x^2 + x + 1}$ and say that the bounds on k are determined by the discriminant.

The discriminant of what? The discriminant of the number k? Please define the discriminant of a number.

The discriminant of the rational expression $\dfrac{x^2 - x + 1}{x^2 + x + 1}?$. Please define the discriminant of a rational expression.

The discriminant of the equation as a whole? Please define the discriminant of an equation.

Perhaps you mean the discriminant of a quadratic expression in one variable. In the case of your rational expression $\dfrac{x^2 - x + 1}{x^2 + x + 1}$, which quadratic expression?

I have to guess what you are talking about. I think you are asking about two propositions that are close to this.

$x^2 + x * f(y) + g(y) = 0 \implies x \in \mathbb\ R\ if\ and\ only\ if\ 0 \le \{f(y)\}^2 - 4 * g(y).$

That proposition is inarguably true, but it is a mere corollary to the fact that the discriminant of the quadratic expression involved in a quadratic equation in standard form determines whether there are real roots or not, which seems to me be to be the fundamental proposition.

The other proposition that I think you are trying to express is this:

If a system of n - 1 equations in n real variables can be reduced to an equation of the form $x^2 + x * f(y) + g(y) = 0$ and if it is possible to solve the inequality

\$0 \le \{f(y)\}^2 - 4 * g(y), then bounds can be set on the possible values of y, and there are real values of x and y satisfying the system.

That too is a true statement, but again my personal opinion is that it is not fundamental. It too is a corollary of the proposition that the discriminant of a quadratic function determines whether that quadratic has real roots.

Now perhaps I have missed the mark because I do not understand what proposition or propositions your examples are designed to demonstrate. I suggest that the next step is for you to explicitly state what your propositions are.

13. Re: A fundamental question

Hi JeffM,

Yes, there has been some lack of clarity on the propositions. And it was your pointing out that I made a thorough search and found this book titled ' Theory of Equations' by Burnside and Panton. This concept is given in that book. The concept is basically of the range of a real valued function and the method used is Transformation or more specifically Homographic Transformation. By expressing x in terms of k we are actually transforming the original expression into a resultant equation which gives us the range of the original expression. It was your direction that led me to search and find the concept.

When I was in school I used to feel that maths theory sucks, most of the students do. Most of the students in our country are good solvers but very few actually know what they are doing. But now I realize theory's worth. As I did my CA I did a lot of business math that involved calculus. I could solve most of those business calculus problems, most of the CAs can, but believe me very few actually know what calculus is. Recently, I came along this calculus book title calculus by Dummies. It's amazing. For the first time I realize what limits are, what a derivative is etc. I realize that calculus has unlimited applications in finance.

So in order to understand theory better I am looking for a E-book titled 'The hitchhiker's guide to calculus' by Michael Spivak. If you or anyone could share that book with me or could tell me where I can download that book from the Net for free, I will be grateful. I am saving bucks for later more useful books. Please inform.

Thanks!