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If the ratio of the roots of $ax^2+bx+c=0$ is equal to the ratio of the roots of $px^2+qx+r=0$ and if $D_1,\:D_2$ are their discriminants respectively, then $D_1:D_2=?$

$\begin{array}{1 1} a^2:p^2 \\ b^2 :q^2 \\ c^2:r^2 \\ b:p \end{array} $

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  • $(a-b)^2=(a+b)^2-4ab$
Let $\alpha_1,\:\beta_1$ are roots of $ax^2+bx+c=0$ and
$\alpha_2,\:\beta_2$ are roots of $px^2+qx+r=0$
Given:$\large\frac{\alpha_1}{\beta_1}$$=\large\frac{\alpha_2}{\beta_2}$ and $D_1=b^2-4ac,\:\:D_2=q^2-4pr$
Also we know that
$\alpha_1+\beta_1=\large-\frac{b}{a}$, $\alpha_1\beta_1=\large\frac{c}{a}$
$\alpha_2+\beta_2=\large-\frac{q}{p}$, $\alpha_2\beta_2=\large\frac{r}{p}$
Using componendo and dividendo we get
$\large\frac{\alpha_1+\beta_1}{\alpha_1-\beta_1}=\frac{\alpha_2+\beta_2}{\alpha_2-\beta_2}$
$\Rightarrow\:\large\frac{(\alpha_1+\beta_1)^2}{(\alpha_1-\beta_1)^2}=\frac{(\alpha_2+\beta_2)^2}{(\alpha_2-\beta_2)^2}$
$\Rightarrow\:\Large\frac{\Large\frac{b^2}{a^2}}{\Large\frac{b^2}{a^2}-4\Large\frac{c}{a}}$=$\Large\frac{\Large\frac{q^2}{p^2}}{\Large\frac{q^2}{p^2}-4\frac{r}{p}}$
$\Rightarrow\:\large\frac{b^2}{b^2-4ac}=\frac{q^2}{q^2-4pr}$
$\Rightarrow\:\large\frac{b^2}{D_1}=\frac{q^2}{D_2}$
$\Rightarrow\:D_1:D_2=b^2:q^2$
answered Jul 14, 2013 by rvidyagovindarajan_1
 

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