Let $\alpha$ and $\beta$ denote the roots in $x$ of the equation $m^2(x^2-x)+2mx+3=0$. If $m_1$ and $m_2$ are the two values of $m$ such that $\alpha$ and $\beta$ satisfy the condition $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{3}$, then find the value of $\frac{m_1^2}{m_2}-\frac{

Question

Let $\alpha$ and $\beta$ denote the roots in $x$ of the equation $m^2(x^2-x)+2mx+3=0$. If $m_1$ and $m_2$ are the two values of $m$ such that $\alpha$ and $\beta$ satisfy the condition $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{3}$, then find the value of $\frac{m_1^2}{m_2}-\frac{m_2^2}{m_1}$.

Accepted Answer

-68/3. From the quadratic, the sum and product of roots are functions of $m$. The condition $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{3}$ simplifies to $3(\alpha^2+\beta^2)=4\alpha\beta$, which can be rewritten using $(\alpha+\beta)^2$. Substituting the Vieta expressions gives two values of $m$, and evaluating the required expression yields $-68/3$.

Solution

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