# Algeb

• May 17th 2012, 06:44 PM
srirahulan
Algeb
$\displaystyle a_1b^2_2c_1=a_2b^2_1c_2$This is given.and then If the roots of $\displaystyle a_1x^2+b_1x+c_1=0$ are $\displaystyle \alpha_1,\beta_1$and the roots of $\displaystyle a_2x^2+b_2x+c_2=0$are$\displaystyle \alpha_2,\beta_2$Then prove$\displaystyle \frac {\alpha_1} {\beta_1}=\frac {\alpha_2}{\beta_2}\ \mbox{or}\ \frac {\alpha_1} {\beta_1}=\frac{\beta_2}{\alpha_2}$
• May 18th 2012, 05:59 AM
Sylvia104
Re: Algeb
$\displaystyle a_1b_2^2c_1=a_2b_1^2c_2$ $\displaystyle \implies$ $\displaystyle \frac{b_1^2}{a_1c_1}=\frac{b_2^2}{a_2c_2}.$

Now

$\displaystyle \left(\alpha_1+\beta_1\right)^2 = \alpha_1^2+\beta_1^2+2\alpha_1\beta_1 = \frac{b_1^2}{a_1^2}.$

$\displaystyle \alpha_1\beta_1 = \frac{c_1}{a_1}.$

$\displaystyle \therefore\ \frac{b_1^2}{a_1c_1} = \frac{\alpha_1^2 + \beta_1^2 + 2\alpha_1\beta_1}{\alpha_1\beta_1}.$

Similarly $\displaystyle \frac{b_2^2}{a_2c_2} = \frac{\alpha_2^2 + \beta_2^2 + 2\alpha_2\beta_2}{\alpha_2\beta_2}.$

$\displaystyle \therefore\ \frac{\alpha_1^2+\beta_1^2+2\alpha_1\beta_1}{\alph a_1\beta_1} = \frac{\alpha^2 + \beta^2 + 2\alpha_2\beta_2}{\alpha_2\beta_2}$

$\displaystyle \Rightarrow\ \alpha_1^2\alpha_2\beta_2 + \alpha_2\beta_1^2\beta_2 + 2\alpha_1\beta_1\alpha_2\beta_2$ $\displaystyle =$ $\displaystyle \alpha_1\alpha^2\beta_1 + \alpha_1\beta_1\beta^2 + 2\alpha_1\beta_1\alpha_2\beta_2$

$\displaystyle \Rightarrow\ \alpha_1^2\alpha_2\beta_2 - \alpha_1\beta_1\beta^2 - \alpha_1\alpha^2\beta_1 + \alpha_2\beta_1^2\beta_2 = 0$

$\displaystyle \Rightarrow\ \left( \alpha_1\beta_2 - \alpha_2\beta_1 \right) \left( \alpha_1\alpha_2 - \beta_1\beta_2 \right) = 0$