If alpha, beta, where alpha

Question

(A) Given the quadratic equation $\lambda x^{2} - (\lambda + 3)x + 3 = 0$. From the relation between roots and coefficients, we have $\alpha + \beta =

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Given the quadratic equation $\lambda x^{2} - (\lambda + 3)x + 3 = 0$. From the relation between roots and coefficients, we have $\alpha + \beta = \frac{\lambda + 3}{\lambda}$ and $\alpha \beta = \frac{3}{\lambda}$. Given $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$, which implies $\frac{\beta - \alpha}{\alpha \beta} = \frac{1}{3}$. Since $\alpha  0$. Thus, $\beta - \alpha = \frac{\alpha \beta}{3} = \frac{3/\lambda}{3} = \frac{1}{\lambda}$. Using the identity $(\beta - \alpha)^{2} = (\alpha + \beta)^{2} - 4\alpha \beta$, we get: $(\frac{1}{\lambda})^{2} = (\frac{\lambda + 3}{\lambda})^{2} - 4(\frac{3}{\lambda})$. $\frac{1}{\lambda^{2}} = \frac{\lambda^{2} + 6\lambda + 9}{\lambda^{2}} - \frac{12}{\lambda}$. Multiplying by $\lambda^{2}$ (where $\lambda 
eq 0$): $1 = \lambda^{2} + 6\lambda + 9 - 12\lambda$. $\lambda^{2} - 6\lambda + 8 = 0$. $(\lambda - 2)(\lambda - 4) = 0$. So, $\lambda = 2$ or $\lambda = 4$. The sum of all possible values of $\lambda$ is $2 + 4 = 6$.

If $\alpha, \beta$, where $\alpha < \beta$, are the roots of the equation $\lambda x^{2} - (\lambda + 3)x + 3 = 0$ such that $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$, then the sum of all possible values of $\lambda$ is:

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