How to make correct substitutions?

Question

If the sum of the roots of quadratic equation, $ax^2+bx+c=0$ is $12$ then the sum of the roots of the equation $a(x+1)^2+b(x+1)+c=0$ is?

I can solve this by procedural way using sum of roots concept, but when I saw the solution in book they wrote:

$\alpha+\beta=12$, where $\alpha$, $\beta$ are the roots of first equation.

Now, let $x=\alpha-1\implies\alpha=x+1$,

replace $x$ by $x=1$

$$a(x+1)^2+b(x+1)+c=0$$ $\implies$ Roots are $\alpha-1, \beta-1$.

Now sum: $\alpha-1+\beta-1=10$

My question is why and how did they substituted $x=\alpha-1$, is it allowed to fix variability of equation by known, and if we substituted something like $\alpha-a=x$ where $a\neq 1$ and followed same step, why we'll not get other answer?

I'm not comfortable with these substitution, please help.

Answer 1

The sum of the roots of $ax^2+bx+c=0$ is $x_0+x_1=12$.

Let $x=x'+1$. Then the sum of the roots of $$ax^2+bx+c=a(x'+1)^2+b(x'+1)+c=0$$ is $x_0+x_1=12=x'_0+1+x'_1+1$ and $$x'_0+x'_1=10.$$

Answer 2

One more:

Rewriting the first equation with the variable y:

$ay^2 +by + c = 0$, $ y_1 + y_2 = 12$.

Note: Setting $y = x+ 1$ in the first equation gives the second equation.

Hence:

$y_1 + y_2 = (x_1 + 1) + ( x_2 + 1) = 12$;

$x_1 + x_2 =10$, where

$x_1,x_2$ are the roots of

$a(x+1)^2 + b(x+1) + c =0$.