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If $\alpha, \beta$ arethe roots of $a x^{2}+2 b x+c=0$ and $\alpha+\delta, \beta+\delta$ are the roots of $\mathrm{Ax}^{2}+2 \mathrm{Bx}+\mathrm{C}=0$, then what is $\left(\mathrm{b}^{2}-\mathrm{ac}\right) / \mathrm{B}^{2}-\mathrm{AC}$
equal to?
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equal to?
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Verified Answer
The correct answer is:
$(a / \mathrm{A})^{2}$
Since, $\alpha$ and $\beta$ are the roots of $a x^{2}+2 b x+c=0$
so, $k+\beta=-\frac{2 b}{a}$ and $\alpha \beta=\frac{c}{a}$
$\delta$ and $\beta+\delta$ are theroots of $A x^{2}+2 B x+C=0$
so, sum of the roots $=\alpha+\beta+2 \delta=-\frac{2 \mathrm{~B}}{\mathrm{~A}}$ and product
ofthe roots $(\alpha+\delta)(\beta+\delta)=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow-\frac{2 b}{a}+2 \delta=-\frac{2 B}{A}$
$\Rightarrow \delta=\frac{b}{a}-\frac{B}{A}$.. (i)
and $(\alpha+\delta)(\beta+\delta)=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \alpha \beta+(\alpha+\beta) \delta+\delta^{2}=\frac{\mathrm{C}}{\mathrm{A}}$...(ii)
Putting value of $\delta$ from equation (i) in equation (ii),
$\frac{\mathrm{c}}{\mathrm{a}}-\frac{2 \mathrm{~b}}{\mathrm{a}}\left(\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{B}}{\mathrm{A}}\right)+\left(\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}-\frac{2 \mathrm{~b}^{2}}{\mathrm{a}^{2}}+\frac{2 \mathrm{bB}}{\mathrm{aA}}+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}+\left(\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}-\frac{2 \mathrm{bB}}{\mathrm{aA}}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}-\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}+\left(\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{B}^{2}}{\mathrm{~A}^{2}}-\frac{\mathrm{C}}{\mathrm{A}}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{\mathrm{B}^{2}-\mathrm{AC}}{\mathrm{A}^{2}}=\frac{\mathrm{b}^{2}-\mathrm{ac}}{\mathrm{a}^{2}}$
$\Rightarrow \frac{b^{2}-a c}{B^{2}-A C}=\left(\frac{a}{A}\right)^{2}$
so, $k+\beta=-\frac{2 b}{a}$ and $\alpha \beta=\frac{c}{a}$
$\delta$ and $\beta+\delta$ are theroots of $A x^{2}+2 B x+C=0$
so, sum of the roots $=\alpha+\beta+2 \delta=-\frac{2 \mathrm{~B}}{\mathrm{~A}}$ and product
ofthe roots $(\alpha+\delta)(\beta+\delta)=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow-\frac{2 b}{a}+2 \delta=-\frac{2 B}{A}$
$\Rightarrow \delta=\frac{b}{a}-\frac{B}{A}$.. (i)
and $(\alpha+\delta)(\beta+\delta)=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \alpha \beta+(\alpha+\beta) \delta+\delta^{2}=\frac{\mathrm{C}}{\mathrm{A}}$...(ii)
Putting value of $\delta$ from equation (i) in equation (ii),
$\frac{\mathrm{c}}{\mathrm{a}}-\frac{2 \mathrm{~b}}{\mathrm{a}}\left(\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{B}}{\mathrm{A}}\right)+\left(\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}-\frac{2 \mathrm{~b}^{2}}{\mathrm{a}^{2}}+\frac{2 \mathrm{bB}}{\mathrm{aA}}+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}+\left(\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}-\frac{2 \mathrm{bB}}{\mathrm{aA}}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}-\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}+\left(\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{B}^{2}}{\mathrm{~A}^{2}}-\frac{\mathrm{C}}{\mathrm{A}}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{\mathrm{B}^{2}-\mathrm{AC}}{\mathrm{A}^{2}}=\frac{\mathrm{b}^{2}-\mathrm{ac}}{\mathrm{a}^{2}}$
$\Rightarrow \frac{b^{2}-a c}{B^{2}-A C}=\left(\frac{a}{A}\right)^{2}$
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