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The ratio of roots of $x^2+bx+c=0$ is same as the ratio of the roots of $x^2+qx+r=0$. Then which of the following is true?

$\begin{array}{1 1}(A) \;br^2=qc^2 \\(B)\;cr^2=qb^2\\(C)\;rc^2=bq^2\\(D)\;rb^2=cq^2 \end{array}$

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Toolbox:
  • $(a-b)^2=(a+b)^2-4ab$
Let the roots of $x^2+bx+c=0$ be $\alpha_1\:and\:\beta_1$
and the roots of $x^2+qx+r=0$ be $\alpha_2\:and\:\beta_2$
$\Rightarrow\:\alpha_1+\beta_1=-b$, $ \alpha_1\beta_1=c$ and
$\alpha_2+\beta_2=-q,\:\:\alpha_2\beta_2=r$
Given: The ratio of the roots of
$x^2+bx+c=0\:\:and\:\:x^2+qx+r=0$ are same
$\Rightarrow\:\large\frac{\alpha_1}{\beta_1}=\frac{\alpha_2}{\beta_2}$
using componendo and dividendo
$\large\frac{\alpha_1+\beta_1}{\alpha_1-\beta_1}=\frac{\alpha_2+\beta_2}{\alpha_2-\beta_2}$
Squaring both the sides,
$\Rightarrow\:\bigg(\large\frac{\alpha_1+\beta_1}{\alpha_1-\beta_1}\bigg)^2=\bigg(\frac{\alpha_2+\beta_2}{\alpha_2-\beta_2}\bigg)^2$
$\Rightarrow$$\large\frac{b^2}{(\alpha_1+\beta_1)^2-4\alpha_1\beta_1}=\frac{q^2}{(\alpha_2+\beta_2)^2-4\alpha_2\beta_2}$
$\Rightarrow\:\large\frac{b^2}{b^2-4c}=\frac{q^2}{q^2-4r}$
$\Rightarrow\:b^2q^2-4b^2r=b^2q^2-4q^2c$
$\Rightarrow\:b^2r=q^2c$
answered Jul 26, 2013 by rvidyagovindarajan_1
 

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