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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given $\alpha^2 - 5\alpha + 3 = 0$ and $\beta^2 - 5\beta + 3 = 0$. This 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: $S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = \frac{5^2 - 2(3)}{3} = \frac{25 - 6}{3} = \frac{19}{3}$. Product of new roots: $P = \frac{\alpha}{\beta} 	imes \frac{\beta}{\alpha} = 1$. The required equation is $x^2 - Sx + P = 0$. Substituting the values: $x^2 - \frac{19}{3}x + 1 = 0$. Multiplying by $3$,we get $3x^2 - 19x + 3 = 0$.

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

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