For which values of a,b,c is this true?

[Ted123]

Homework Statement

For which values of a,b,c\in\mathbb{C} is the following equation true? a(x+t)^2 + b(x+t) + c = \alpha(ax^2 + bx + c) where \alpha is some scalar.

The Attempt at a Solution

How do I go about this?

 

[haruspex]

I don't understand the question. What roles do x and t play here? Is the equation to be true for all x and t? Some x and t? ...

 

[Ted123]

I don't understand the question. What roles do x and t play here? Is the equation to be true for all x and t? Some x and t? ...

x is a variable and t\in\mathbb{R} and \alpha is a fixed constant. We want the equation to be true for all t.

If you equate coefficients you get:

a=\alpha a
2ta+b = \alpha b
at^2 + bt + c = \alpha c

For what values of a, b and c are these true?

 

[haruspex]

x is a variable and t\in\mathbb{R} and \alpha is a fixed constant. We want the equation to be true for all t.

You mean for all x, I assume. Is alpha real?

If you equate coefficients you get:

a=\alpha a
2ta+b = \alpha b
at^2 + bt + c = \alpha c

For what values of a, b and c are these true?

Try considering α=1, α≠1 separately. (That's alpha, not a.)

 

[Ted123]

You mean for all x, I assume. Is alpha real?

Try considering α=1, α≠1 separately. (That's alpha, not a.)

If \alpha =1 then a=0 and b=0

If \alpha \neq 1 then the first equation implies a(1-\alpha) = 0 so a=0 since \alpha \neq 1.

Subbing a=0 into the second equation gives b=\alpha b so b(1-\alpha)=0 so b=0 since \alpha \neq 1.

Subbing a=0, b=0 in the third equation gives c=0

 

[haruspex]

If \alpha =1 then a=0 and b=0

Unless t = 0.

 

[Ted123]

Unless t = 0.

So, assuming t\neq 0 (for if t=0 the equation is trivially true), I can conclude that the equation will be true for \alpha =1 for all c\in\mathbb{C}, a=0, b=0 and when \alpha \neq 1 it will only be true for a,b,c=0?

In other words, whatever the value of \alpha, the equation will be true for all c\in\mathbb{C} with a,b=0 so the polynomial involved p(x) = ax^2 + bx + c must be constant; i.e. p(x) = c.