Algeb

**srirahulan** · May 17th 2012, 06:44 PM

$a_1b^2_2c_1=a_2b^2_1c_2$ This is given.and then If the roots of $a_1x^2+b_1x+c_1=0$ are $\alpha_1,\beta_1$ and the roots of $a_2x^2+b_2x+c_2=0$ are $\alpha_2,\beta_2$ Then prove $\frac {\alpha_1} {\beta_1}=\frac {\alpha_2}{\beta_2}\ \mbox{or}\ \frac {\alpha_1} {\beta_1}=\frac{\beta_2}{\alpha_2}$

**Sylvia104** · May 18th 2012, 05:59 AM

$a_1b_2^2c_1=a_2b_1^2c_2$ $\implies$ $\frac{b_1^2}{a_1c_1}=\frac{b_2^2}{a_2c_2}.$

Now

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

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

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

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

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

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

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

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

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