If alpha ne beta but alpha^2 = 5alpha - 3 and | vedclass

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(D) Given that $\alpha^2 - 5\alpha + 3 = 0$ $(i)$ and $\beta^2 - 5\beta + 3 = 0$ $(ii)$.This implies that $\alpha$ and $\beta$ are the roots of the qu

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

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(D) Given that $\alpha^2 - 5\alpha + 3 = 0$ $(i)$ and $\beta^2 - 5\beta + 3 = 0$ $(ii)$.This implies that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 5x + 3 = 0$.From the properties of roots,the sum of roots is $\alpha + \beta = 5$ and the product of roots is $\alpha\beta = 3$.We need to find the equation whose roots are $p = \alpha/\beta$ and $q = \beta/\alpha$.The sum of the new roots is $p + q = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$.Since $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 5^2 - 2(3) = 25 - 6 = 19$,we have $p + q = \frac{19}{3}$.The product of the new roots is $pq = \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$.The required quadratic equation is $x^2 - (p + q)x + pq = 0$.Substituting the values,we get $x^2 - \frac{19}{3}x + 1 = 0$.Multiplying by $3$,we obtain $3x^2 - 19x + 3 = 0$.

If $\alpha \ne \beta$ but $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$,then the equation whose roots are $\alpha/\beta$ and $\beta/\alpha$ is

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