Alternate way to solve this algebra problem?

Question

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?

Answer 1

Note that the difference of roots is actually $$\frac{\sqrt{b^2-4ac}}{|a|}$$ If the roots be $p$ and $q$, then $$|p-q| =\sqrt{(p-q)^2} = \sqrt{(p+q)^2 -4pq} = \sqrt{\frac{b^2}{a^2} -4\frac ca}$$ (using Vieta’s)

Answer 2

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, $| y_1-y_2|= 12$.

Answer 3

Alternative approach

Since $r_1, r_2$ are the roots of $ax^2 + bx + c = 0$,

they are also the roots of $(36a)x^2 + (36b)x + (36c) = 0.$

Since (for example) $[6r_1]^2 = 36(r_1)^2~$ and $~6(6r_1) = 36r_1$ then

$(6 r_1), (6 r_2)$

are the roots of $ax^2 + 6bx + 36c = 0$.

Since $|r_1 - r_2| = 2, ~|(6r_1) - (6 r_2)| = 12.$