# Theory of cubic equation question

1. Dec 1, 2012

### lionely

1. The problem statement, all variables and given/known data
Find the equation who roots are the squares of the roots of the equation x^3-4x^2+x-

x^3-16y^2+y+1?

2. Dec 1, 2012

### SammyS

Staff Emeritus
Neither of these is an equation.

An equation has an equal sign with a mathematical expression on each side of the equal sign..

3. Dec 1, 2012

### lionely

omg i didn't notice my keyboard has a problem ..........................

it's x^3-4x^2+x-1=0

x^3-16x^2+x+1=0

Last edited: Dec 1, 2012
4. Dec 1, 2012

### Michael Redei

That second expression isn't an equation. But have you actually worked out what steps you need to take to solve the problem? Here's a hint:

Equations can have roots.
Roots have squares.
These squares can be roots of other equations.

5. Dec 1, 2012

### lionely

I know i took the roots of the 1st equation to be x so the ones for the new equation are x^2.

I said let y= new roots, y= x^2
x= sqrt y

and I put x in first equation and I got x^3-16x^2+x+1=0

Last edited: Dec 1, 2012
6. Dec 1, 2012

### Dick

Can you show all of what you did? I think that's way oversimplified. This is a problem involving symmetric polynomials, isn't it?

7. Dec 1, 2012

### lionely

x^3-4x^2+x-1=0

new roots = x ^2
let y=x^2, x = sqrt y

(sqrty)^3-4(sqrty)^2+sqrty - 1
(y^3/2 - 4y + sqrty -1)=0
squaring both sides
y^3+16y^2+y+1

8. Dec 1, 2012

### Dick

No, that is not the square. That's just bad algebra. E.g. (sqrty-1)^2 isn't y+1.

9. Dec 1, 2012

### lionely

D:, I'll leave it as (y^3/2 - 4y + sqrty -1)=0 then.

10. Dec 1, 2012

### Dick

I think you missed the lecture were they were actually talking about how to do this problem. Sure, you could do that, but it's not a polynomial and it's not particularly well defined either if all of the roots aren't positive. What you missed is that if your polynomial has three roots a,b,c then you can write it as (x-a)(x-b)(x-c) then the coefficients of your original polynomial are various symmetric polynomials of a,b and c. The polynomial you want is (x-a^2)(x-b^2)(x-c^2). The coefficients are also symmetric polynomials of a, b, and c. Doesn't that ring some kind of bell?

11. Dec 1, 2012

### lionely

oh.. Oh Oh Let me do it over

12. Dec 1, 2012

### lionely

Is it a^6+3 a^4 y+7 a^4+3 a^2 y^2+14 a^2 y-5 a^2+y^3+7 y^2-5 y-1?

13. Dec 1, 2012

### Dick

Not likely. Why do you even think it would be that? Could you start giving reasons why you think the answers you post are true?

14. Dec 1, 2012

### lionely

Isn't it

y= x-a^2

so x = y+a^2? then i put it into the equation?

15. Dec 2, 2012

### Dick

That makes no sense. Expand (x-a)(x-b)(x-c)=0 and compare it with x^3-4x^2+x-1=0. You can express the numbers in the second polynomial as functions of a,b,c. What are they? You really haven't talked about this kind of thing in class or in the textbook?

16. Dec 2, 2012

### SammyS

Staff Emeritus
@lionely,

Look at the following post from Dick again !

Multiplying (x-a)(x-b)(x-c) out gives: $\displaystyle x^3 - (a+b+c)x^2+(ab+bc+ca)x-abc\ .$

Equate this and $\displaystyle x^3-4x^2+x-1\ .$

You will not be able to solve for the a, b , or c individually, but you can find $\displaystyle (a+b+c)\,,\ (ab+bc+ca)\,,\ \$ and $\ \ \displaystyle (abc)\ \$ by inspection.

Now multiply out (x-a2)(x-b2)(x-c2) . With a bit of algebra and some cleverness, you will be able find your desired polynomial.

17. Dec 2, 2012

### micromass

18. Dec 2, 2012

### lionely

Is it √x(x+1) = 1 + 4x ?

19. Dec 2, 2012

### Dick

No. Go back and read the previous posts and think about them. If you don't want to do that, then forget this problem and move on to the next one.

20. Dec 2, 2012

### Curious3141

Your method is basically sound, but I'm betting you screwed up the algebra. I did it your way and got the right answer. You have to be very careful in grouping terms and squaring out the surd forms.

While others have suggested using a slightly more involved method to compare the sums/products of roots to the coefficients, this is unnecessary here.