# Crafting an equation using roots of another equation

1. Jul 30, 2016

### Faiq

1. The problem statement, all variables and given/known data
Find the equation whose roots are given by adding 2 to the roots of the equation
x^4+3x^3-13x^2-51x-36=0

2. Relevant equations
x^4-(Σa)x³+(Σab)x²-(Σabc)x+abcd = 0

3. The attempt at a solution

And the coefficient of x^3 to be -5
second coefficient = b/a = -Σa = 3
Since we are adding two to every root so b/a = -(Σa +2) = -(Σa +Σ2) = -(-3+8) = -5 thus the equation should have x^4-5x³

And I am stuck at x^2
=Σ(a+2)(b+2)
=Σab+2a+2b+4
=Σab+ 4Σa +16 which amounts to -9
Correct coefficient is -7

Note:- We aren't allowed to solve the given equation in any part of the working

2. Jul 30, 2016

### LCKurtz

Hint: If $r$ is a root of $f(x)$ what would be a root of $f(x-2)$?

3. Jul 30, 2016

### ehild

It is wrong. Show your working.
The correct coefficient is really -7.

4. Jul 31, 2016

### Faiq

r+2

5. Jul 31, 2016

### Faiq

Ehild, I am aware the coefficient is wrong, the working I have given is my wrong working and I want to know where I went wrong

6. Jul 31, 2016

### haruspex

Right. Can you see how to use that to make a substitution for x in the polynomial?

7. Jul 31, 2016

### Faiq

I am aware of the method you are implying, but the question forbids me to use that method.

8. Jul 31, 2016

### haruspex

No, it says
which is not what I have in mind.

9. Jul 31, 2016

### ehild

Your notations are not clear. What do you mean with Σa?
If x1, x2, x3 x4 are the roots of the original equation, the coefficient of the x2term is
Σ(xi+2)(xj+2) for all pairs i<j: it is 6 terms as you have 6 pairs from 4 elements.
So it is (x1x2+x1x3+x1x4+x2x3+x2x4+x3x4)+2(x1+x2+x1+x3+x1+x4+x2+x3+x2+x4+x3+x4)+6*4, as you have 6 times 2*2.
Also you have 3Σxi in the parentheses of the second term, instead of 2.

Last edited: Jul 31, 2016