# algebra fix

• September 8th, 2014, 06:05 AM
parag29081973
algebra fix
I want to know the results of following equations.
x2 + x +1 = y
In the above equation for 1 value of x there is 1 value of y. But for 1 value of y there are three values of x as it then becomes a 3rd degree equation.
1.If we go on substituting values for the variable x we get 1 value of y.
2.If we substitute values for y we get 3 values for x.
Aren't case 1 and case 2 supposed to yield the same result?
• September 8th, 2014, 08:42 AM
river_rat
Quote:

Originally Posted by parag29081973
I want to know the results of following equations.
x2 + x +1 = y
In the above equation for 1 value of x there is 1 value of y. But for 1 value of y there are three values of x as it then becomes a 3rd degree equation.
1.If we go on substituting values for the variable x we get 1 value of y.
2.If we substitute values for y we get 3 values for x.
Aren't case 1 and case 2 supposed to yield the same result?

Um, that is a 2nd degree equation?
• September 8th, 2014, 04:06 PM
mathman
Quote:

Originally Posted by parag29081973
I want to know the results of following equations.
x2 + x +1 = y
In the above equation for 1 value of x there is 1 value of y. But for 1 value of y there are three values of x as it then becomes a 3rd degree equation.
1.If we go on substituting values for the variable x we get 1 value of y.
2.If we substitute values for y we get 3 values for x.
Aren't case 1 and case 2 supposed to yield the same result?

As the previous comment noted it is quadratic in x, not cubic. In general functions don't have to be 1-1. y = x^2 is even simpler.
• September 11th, 2014, 02:59 PM
Guitarist
True.

It might just be worth pointing out that it usual to make a distinction between polynomial forms i.e. expressions like

(where the are scalar and the are drawn from any commutative ring and the subscripts on the former are indices and the superscripts on the latter are exponents)

and polynomial functions which have the same form but a perfectly well-defined domain and co-domain.

Example the distinct forms and

each define the same function (the identity ) when , the ring of integers mod 2 but not (necessarily) otherwise

It may also be worth pointing that , if we take the highest exponent with non-zero coefficient to be the degree of a polynomial form (or function - doesn't matter which) then any degree polynomial as exactly values of that satisfy . These are called the "roots" or "zeros" of a polynomial form or function. This is known as the Fundamental Theorem of Algebra. (proof non-trivial!)

The polynomial in the OP accordingly has 2 roots. Try to find them if you dare, but I don't advise it - you'll freak!!
• September 12th, 2014, 06:31 AM
Laurieag
Quote:

Originally Posted by Guitarist
The polynomial in the OP accordingly has 2 roots. Try to find them if you dare, but I don't advise it - you'll freak!!

That's interesting Guitarist, in an imaginary way.

In the standard quadratic form the equation is , so a equals 1, b equals 1 and c equals 1.

The quadratic roots can be obtained by solving with a, b and c all equal to 1.
• September 12th, 2014, 02:13 PM
Guitarist
Quote:

Originally Posted by Laurieag
That's interesting Guitarist, in an imaginary way.

Ha ha - I assume you are referring to something like the following.......

Using a recipe related to yours, but slightly more general, I got that the 2 roots of are

which (unless I messed up a sign somewhere - my usual failing) works quite nicely Note the imaginary unit, which leads me to the following......

A ring element that is the root of some polynomial (with rational coefficients) is called an "algebraic number", otherwise it is called a "transcendental number". Algebraic numbers are, in general, complex numbers - I go further: if an algebraic number does not "contain" the imaginary unit, we simply note that the Real numbers are a sub-set of the Complex numbers and Bingo!

You may not like the idea that the monster above is a number, but tough - you can see it is if you do the calculation (to prove it is the root of the OP polynomial)
• September 13th, 2014, 02:33 AM
Laurieag
Quote:

Originally Posted by Guitarist
Using a recipe related to yours, but slightly more general, I got that the 2 roots of are

which (unless I messed up a sign somewhere - my usual failing) works quite nicely

Hi Guitarist,

I get a slightly different answer of and although they are equivalent if you mean .

first root = and the second =

Further, in the equation , the square of each of the roots gives the other root and both roots multiplied together gives 1 (If I've got the signs straight).

= , = and

Whoever coined the term 'imaginary unit' certainly had a good imagination.
• September 13th, 2014, 11:32 AM
someguy1
Quote:

Originally Posted by Laurieag

... and

both roots multiplied together gives 1 (If I've got the signs straight).

Surely that can't be right in general, what is the product of the roots of x^2 - 4 = 0? You dropped out c/a.

In any event a good way to look at this equation is that if x^3 = 1 then

x^3 - 1 = (x-1)(x^2 + x + 1) = 0

One root is x = 1, and the other two are the roots of the second factor, the subject of the thread. So these numbers are the complex cube roots of 1. If you graph them on the plane you'll see why the square of each of the non-real roots is the other one. And you'll also see why the real and imaginary parts of the roots are the sine and cosine values associated with the 30-60-90 triangle except for the signs.
• September 13th, 2014, 03:05 PM
Robittybob1
Quote:

Originally Posted by parag29081973
I want to know the results of following equations.
x2 + x +1 = y
In the above equation for 1 value of x there is 1 value of y. But for 1 value of y there are three values of x as it then becomes a 3rd degree equation.
1.If we go on substituting values for the variable x we get 1 value of y.
2.If we substitute values for y we get 3 values for x.
Aren't case 1 and case 2 supposed to yield the same result?

The answer to a question could be yes or no, but there are endless questions that will have yes as the answer. Your logic is being questioned. Yes or No?
• September 14th, 2014, 05:50 PM
Laurieag
Quote:

Originally Posted by someguy1
Surely that can't be right in general, what is the product of the roots of x^2 - 4 = 0?

That's why I specified the equation given in the OP. a, b and c equal 1 as well as both roots multiplied by each other.

Quote:

Further, in the equation