If alpha neq beta but alpha^2 = 5alpha - 3 an | vedclass

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(D) Given $\alpha^2 - 5\alpha + 3 = 0$ and $\beta^2 - 5\beta + 3 = 0$,$\alpha$ and $\beta$ are roots of the equation $x^2 - 5x + 3 = 0$.From the prope

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$3x^2 - 19x + 3 = 0$

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(D) Given $\alpha^2 - 5\alpha + 3 = 0$ and $\beta^2 - 5\beta + 3 = 0$,$\alpha$ and $\beta$ are roots of the equation $x^2 - 5x + 3 = 0$.From the properties of roots,$\alpha + \beta = 5$ and $\alpha\beta = 3$.We need to find the equation with roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.Sum of roots $= \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{5^2 - 2(3)}{3} = \frac{25 - 6}{3} = \frac{19}{3}$.Product of roots $= \frac{\alpha}{\beta} 	imes \frac{\beta}{\alpha} = 1$.The required equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - \frac{19}{3}x + 1 = 0$.Multiplying by $3$,we get $3x^2 - 19x + 3 = 0$.

If $\alpha \neq \beta$ but $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$,then the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is:

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