<s>[INST] 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.
 [/INST] 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.
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