Proving a relation related to quadratic equation

Question

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,

$α+β=\frac{-2b}{a}$

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$αβ=\frac{c}{a}$

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$α+δ+β+δ=\frac{-2B}{A}$

⇒ $α+β+2δ =\frac{-2B}{A}$

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$(α+δ)(β+δ)=\frac{C}{A}$

⇒ $αβ+αδ+βδ+δ^2=\frac{C}{A}$

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

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$(α+β)^2=\frac{4b^2}{a^2}$

⇒ $α^2+β^2+2αβ=\frac{4b^2}{a^2}$

$α^2+β^2+\frac{2c}{a}=\frac{4b^2}{a^2}$

⇒ $α^2+β^2=\frac{4b^2-2ac}{a^2}$ -(1)

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$(α+β+2δ)^2 =\frac{4B^2}{A^2}$

⇒ $α^2+β^2+(2δ)^2+2(αβ+2βδ+2αδ)=\frac{4B^2}{A^2}$

⇒$α^2+β^2+4δ^2+2αβ+4βδ+4αδ=\frac{4B^2}{A^2}$

⇒$α^2+β^2+2αβ+4(δ^2+βδ+αδ)=\frac{4B^2}{A^2}$

⇒$α^2+β^2+\frac{2c}{a}+4(\frac{Ca-cA}{Aa})=\frac{4B^2}{A^2}$

⇒$α^2+β^2=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$ -(2)

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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.

Answer 1

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:

$$\require{cancel} B^2 - C=(b^2-\bcancel{2 b\delta}+\cancel{\delta^2})-(\cancel{\delta^2}-\bcancel{2b\delta}+c) = b^2 - c $$

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[ 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): $$\require{cancel} \frac{4b^2-2ac}{a^\bcancel{2}}=\frac{4aB^2-2A^2 c-4Aa\color{red}{C}+4cA^2}{A^2 \bcancel{a}} $$

$$ 4b^2A^2-\bcancel{2acA^2}=4a^2B^2-\bcancel{2acA^2}-4a^2AC+4acA^2 $$

$$ \bcancel{4}A^2(b^2-ac) = \bcancel{4}a^2(B^2-AC) $$

$$\frac{b^2-ac}{a^2} =\frac{B^2-AC}{A^2} $$

Answer 2

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 :

$$|\alpha-\beta|=|(\alpha+\delta)-(\beta+\delta)|=\sqrt{(\alpha+\beta)^2-4\alpha \beta}=\sqrt{(\alpha+\delta +\beta+ \delta)^2-4(\alpha+\delta)( \beta+\delta)}$$

$$\implies \sqrt{\left(\frac {-2b}{a} \right)^2 -4\left(\frac ca \right)}= \sqrt{\left(\frac {-2B}{A} \right)^2 -4\left(\frac CA\right)}$$

$$\implies \frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$$

Answer 3

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

$$b^2-c=(\delta+B)^2-(\delta^2+2B\delta+C)=B^2-C.$$

Answer 4

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.

Answer 5

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.