Crafting an equation using roots of another equation

[Faiq]

Homework Statement

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

Homework Equations

x^4-(Σa)x³+(Σab)x²-(Σabc)x+abcd = 0

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

x^4-(Σa)x³+(Σab)x²-(Σabc)x+abcd = 0

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

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

 

[ehild]

Homework Statement

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

Homework Equations

x^4-(Σa)x³+(Σab)x²-(Σabc)x+abcd = 0

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

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

Correct coefficient is -7

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

 

[Faiq]

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

r+2

 

[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

 

[haruspex]

r+2

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

 

[Faiq]

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

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

 

[haruspex]

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

No, it says

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

which is not what I have in mind.

 

[ehild]

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

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 3Σx_i in the parentheses of the second term, instead of 2.