Alternate way to solve this algebra problem? Working on a problem that says given the absolute value of the difference of the roots of $ax^2 + bx + c$ as $2$, what is the absolute value of the difference of the roots of $ax^2 + 6bx + 36c$?
I reasoned that the difference of roots would just be $2\sqrt{b^2 - 4ac}$. Therefore, the value of $\sqrt{b^2 - 4ac}$ is $1$. Then, I used this information to answer the question:
$2\sqrt{36b^2 - 36(4ac)} =$ ?
$2\sqrt{36(b^2 - 4ac)}$
$12\sqrt{b^2 - 4ac}$
$=12$
The question hinted that I could use Vieta's formulas to solve the problem. Is my method correct, and is there a method involving Vieta's formulas?

Note, for $ax^2+bx +c=0$
$$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=\frac{b^2}{a^2}-\frac{4c}a=2^2$$
Then, for $ay^2 + 6by + 36c=0$
$$(y_1-y_2)^2 = (y_1+y_2)^2-4y_1y_2=36\left(\frac{b^2}{a^2}-\frac{4c}a\right)=36\cdot 4
$$
Thus, $| y_1-y_2|= 12$.
