When does a(x+t)² + b(x+t) + c = α(ax² + bx + c)?

[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. $p(x) = c$.