Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$   

$αβ=\frac{c}{a}$

$α+δ+β+δ=\frac{-2B}{A}$ 
⇒ $α+β+2δ =\frac{-2B}{A}$

$(α+δ)(β+δ)=\frac{C}{A}$ 
⇒ $αβ+αδ+βδ+δ^2=\frac{C}{A}$
⇒$\frac{c}{a}+αδ+βδ+δ^2=\frac{C}{A}$ ⇒ $αδ+βδ+δ^2=\frac{Ca-cA}{Aa}$

$(α+β)^2=\frac{4b^2}{a^2}$ 
⇒ $α^2+β^2+2αβ=\frac{4b^2}{a^2}$  
$α^2+β^2+\frac{2c}{a}=\frac{4b^2}{a^2}$
⇒ $α^2+β^2=\frac{4b^2-2ac}{a^2}$ -(1)

$(α+β+2δ)^2 =\frac{4B^2}{A^2}$
⇒ $α^2+β^2+(2δ)^2+2(αβ+2βδ+2αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+4δ^2+2αβ+4βδ+4αδ=\frac{4B^2}{A^2}$
⇒$α^2+β^2+2αβ+4(δ^2+βδ+αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+\frac{2c}{a}+4(\frac{Ca-cA}{Aa})=\frac{4B^2}{A^2}$
⇒$α^2+β^2=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$ -(2)

From (1) and (2) we get,
$\frac{4b^2-2ac}{a^2}=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$
My problem: I tried simplifying it further but could not reach the required result. A continuation of my method would be more appreciated compared to other methods.
 A: Alternative hint: we can assume WLOG that $\,a=A=1\,$, since both the roots and the equality to be proved are homogeneous in the respective coefficients.
Then, if $\alpha, \beta$ are the roots of $x^2+2bx+c=0$, the polynomial with roots $\alpha+\delta, \beta+\delta$ is:
$$(x-\delta)^2+2b (x-\delta)+c=0 \;\;\iff\;\; x^2 + 2(b-\delta)x+\delta^2-2b\delta+c=0\,$$
Identifying coefficients gives $B=b-\delta$ and $C=\delta^2-2b\delta+c\,$, then:
$$\require{cancel}
B^2 - C=(b^2-\bcancel{2 b\delta}+\cancel{\delta^2})-(\cancel{\delta^2}-\bcancel{2b\delta}+c) = b^2 - c
$$

[ EDIT ]  To answer OP's edit:


A continuation of my method would be more appreciated compared to other methods.

There is an error/typo in formula (2). Once corrected (in red below):
$$\require{cancel}
\frac{4b^2-2ac}{a^\bcancel{2}}=\frac{4aB^2-2A^2 c-4Aa\color{red}{C}+4cA^2}{A^2 \bcancel{a}}
$$
$$
4b^2A^2-\bcancel{2acA^2}=4a^2B^2-\bcancel{2acA^2}-4a^2AC+4acA^2
$$
$$
\bcancel{4}A^2(b^2-ac) = \bcancel{4}a^2(B^2-AC)
$$
$$\frac{b^2-ac}{a^2} =\frac{B^2-AC}{A^2}
$$
A: WLOG, $A=a=1$ (otherwise you can divide the trinomials by their leading coefficient).
Then
$$(x+\delta)^2+2B(x+\delta)+C=x^2+2(\delta+B)x+\delta^2+2B\delta+C=x^2+2bx+c,$$ and
$$b^2-c=(\delta+B)^2-(\delta^2+2B\delta+C)=B^2-C.$$
A: Instead doing all the hard work you did, you can notice that the difference of roots $(\vert x_1-x_2\vert )$ is same for both the equations. Hence :
$$|\alpha-\beta|=|(\alpha+\delta)-(\beta+\delta)|=\sqrt{(\alpha+\beta)^2-4\alpha \beta}=\sqrt{(\alpha+\delta +\beta+ \delta)^2-4(\alpha+\delta)( \beta+\delta)}$$
$$\implies \sqrt{\left(\frac {-2b}{a} \right)^2 -4\left(\frac ca \right)}= \sqrt{\left(\frac {-2B}{A} \right)^2 -4\left(\frac CA\right)}$$
$$\implies 
\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$$
A: Here is the LHS:
$$\frac{b^2-ac}{a^2} = \left(\frac{b}{a}\right)^2-\frac {c}{a}$$
$$=\left(\frac {-(\alpha+\beta)} {2}\right)^2-\alpha\beta$$
$$=\frac {{\alpha}^2+{\beta}^2+2\alpha\beta}{4}-\frac {4\alpha\beta}{4}$$
$$=\frac{{\alpha}^2+{\beta}^2-2\alpha\beta} {4}$$
$$=\left(\frac{\alpha-\beta} {2}\right)^2$$
And here is the RHS:
$$\frac{B^2-AC}{A^2}=\left(\frac {B}{A}\right)^2-\frac{C}{A}$$
$$=\left(\frac{-(\alpha+\beta+2\delta)}{2}\right)^2-(\alpha+\delta) (\beta+\delta)$$
$$=\frac {{\alpha}^2+{\beta}^2+4{\delta}^2+2\alpha\beta+4\alpha\delta+4\beta\delta} {4}-\frac {4\alpha\beta+4\alpha\delta+4\beta\delta} {4}$$
$$=\frac {{\alpha}^2+{\beta}^2-2\alpha\beta} {4}$$
$$=\left(\frac {\alpha-\beta} {2}\right)^2$$
Clearly, LHS = RHS. QED.
A: By the formula for the roots of a quadratic equation, the squared difference between them is $$\left(\frac{\pm\sqrt{b^2-ac}}{a}\right)^2=\frac{b^2-ac}{a^2}$$ (factor $4$ omitted) and is invariant by translation.
