If alpha neq beta,alpha^2 = 5alpha - 3,and be | vedclass

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(C) Given $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$,it implies that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 5x +

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

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(C) Given $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$,it implies that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 5x + 3 = 0$.From the properties of roots,we have:Sum of roots: $\alpha + \beta = 5$Product of roots: $\alpha \beta = 3$We need to find the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.Sum of new 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 new roots: $\frac{\alpha}{\beta} 	imes \frac{\beta}{\alpha} = 1$.The required quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - \frac{19}{3}x + 1 = 0$.Multiplying by $3$,we get $3x^2 - 19x + 3 = 0$.

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

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