# If alpha and beta be the value of x obtained from the equation m^2(x^2-x)+2mx+3=0 and if m_1 and m2 be the two value of m for which alpha and beta are connected by the relation alpha by beta +beta by alpha =4/3 find the value of m1^2/m2+m2^2/m1?

May 25, 2018

$- \frac{68}{3}$

#### Explanation:

The equation is

${m}^{2} {x}^{2} + \left(2 m - {m}^{2}\right) x + 3 = 0$

This means that for the two roots $\alpha$ and $\beta$ we have

$\alpha + \beta = \frac{{m}^{2} - 2 m}{m} ^ 2 q \quad \text{and} q \quad \alpha \beta = \frac{3}{m} ^ 2$

Thus

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{{\alpha}^{2} + {\beta}^{2}}{\alpha \beta}$
$q \quad q \quad = \frac{{\left(\alpha + \beta\right)}^{2} - 2 \alpha \beta}{\alpha \beta}$
$q \quad q \quad = {\left(\alpha + \beta\right)}^{2} / \left(\alpha \beta\right) - 2$
$q \quad q \quad = {\left({m}^{2} - 2 m\right)}^{2} / {m}^{4} \times {m}^{2} / 3 - 2$
$q \quad q \quad = {\left({m}^{2} - 2 m\right)}^{2} / \left(3 {m}^{2}\right) - 2$
$q \quad q \quad = {\left(m - 2\right)}^{2} / 3 - 2$

(note that $m \ne 0$ as is obvious from the original equation, which leads to a contradiction for $m = 0$)

and so

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{4}{3} \implies$

${\left(m - 2\right)}^{2} / 3 - 2 = \frac{4}{3} \implies$

${\left(m - 2\right)}^{2} = 10 \implies$

${m}^{2} - 4 m - 6 = 0$

(note that we can easily find the two values ${m}_{1}$ and ${m}_{2}$ to be $2 \pm \sqrt{10}$, respectively - but the algebra is easier if we proceed by using properties of the quadratic equation)

${m}_{1} + {m}_{2} = 4 q \quad \text{and} q \quad {m}_{1} {m}_{2} = - 6$
${m}_{1}^{2} / {m}_{2} + + {m}_{2}^{2} / {m}_{1} = \frac{{m}_{1}^{3} + {m}_{2}^{3}}{{m}_{1} {m}_{2}}$
$q \quad q \quad = \frac{{\left({m}_{1} + {m}_{2}\right)}^{3} - 3 {m}_{1} {m}_{2} \left({m}_{1} + {m}_{2}\right)}{{m}_{1} {m}_{2}}$
$q \quad q \quad = {\left({m}_{1} + {m}_{2}\right)}^{3} / \left({m}_{1} {m}_{2}\right) - 3 \left({m}_{1} + {m}_{2}\right)$
$q \quad q \quad = {4}^{3} / \left(- 6\right) - 3 \times 4 = - \frac{64}{6} - 12 = - \frac{68}{3}$