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If $\alpha, \beta$ are the roots of $a x^{2}+b x+c=0(a \neq 0)$ and $\alpha+h, \beta+h$ are the roots of $p x^{2}+q x+r=0$ $(p \neq 0),$ then the ratio of the squares of their discriminants is
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The correct answer is:
$a^{2}: p^{2}$
Given, $\alpha, \beta$ are the roots of $a x^{2}+b x+c=0$ and
$a+h, \beta+h$ are the roots of $p x^{2}+q x+r=0$
$\therefore \quad \alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$
and $\alpha+h+\beta+h=-\frac{q}{p},(\alpha+h)(\beta+h)=\frac{r}{p}$
Now, $\quad(\alpha+h)-(\beta+h)=\alpha-\beta$
$\Rightarrow \quad[(\alpha+h)-(\beta+h)]^{2}=(\alpha-\beta)^{2}$
$\Rightarrow[(\alpha+h)+(\beta+h)]^{2}-4(\alpha+h)(\beta+h)$
$\Rightarrow \quad \frac{q^{2}-4 p r}{p^{2}}=\frac{b^{2}-4 a c}{a^{2}}$
$\therefore$
$\frac{b^{2}-4 a c}{q^{2}-4 p r}=\frac{a^{2}}{p^{2}}$
Hence, the ratio of the square of their discriminants is $a^{2}: p^{2}$
$a+h, \beta+h$ are the roots of $p x^{2}+q x+r=0$
$\therefore \quad \alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$
and $\alpha+h+\beta+h=-\frac{q}{p},(\alpha+h)(\beta+h)=\frac{r}{p}$
Now, $\quad(\alpha+h)-(\beta+h)=\alpha-\beta$
$\Rightarrow \quad[(\alpha+h)-(\beta+h)]^{2}=(\alpha-\beta)^{2}$
$\Rightarrow[(\alpha+h)+(\beta+h)]^{2}-4(\alpha+h)(\beta+h)$
$\Rightarrow \quad \frac{q^{2}-4 p r}{p^{2}}=\frac{b^{2}-4 a c}{a^{2}}$
$\therefore$
$\frac{b^{2}-4 a c}{q^{2}-4 p r}=\frac{a^{2}}{p^{2}}$
Hence, the ratio of the square of their discriminants is $a^{2}: p^{2}$
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