If the equation $2x^2-7x+12=0$ has two roots alpha and beta ,then the value of alpha/beta+beta/alpha is If the equation $2x^2-7x+12 =0$ has two roots $\alpha$ and $\beta$ ,
then the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ is 
note $x=\frac{-7+\sqrt{47}}{4},\frac{-7-\sqrt{47}}{4}$
then 
$$\frac{\frac{-7+\sqrt{47}}{4}}{\frac{-7-\sqrt{47}}{4}}+\frac{\frac{-7-\sqrt{47}}{4}}{\frac{-7+\sqrt{47}}{4}}$$
so 96/2
 A: $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$
since $\alpha+\beta=\frac{7}{2}$ and $\alpha\beta=6$, you can compute the value by substituting.
A: First, note that your quantity can be rewritten as
$\frac{\alpha^2 + \beta^2}{\alpha \beta}$,
and the numerator and denominator here are both symmetric polynomials in $\alpha$ and $\beta$.
On the other hand, for any polynomial, the coefficients themselves are in fact symmetric polynomials in the roots. By dividing by the leading coefficient (which doesn't change the roots), we may as well assume that a polynomial is monic (which in our case gives us $x^2 - \frac{7}{2} x + 6$).
Now, if $\alpha$ and $\beta$ are the roots of a monic polynomial $x^2 + bx + c$, we must have
$x^2 + b x + c = (x - \alpha)(x - \beta) = x^2 + (- (\alpha + \beta)) x + \alpha \beta$, and comparing like coefficients gives
$-b = \alpha + \beta$ and $c = \alpha \beta$.
In fact, these elements generate the ring of symmetric polynomials in $\alpha$ and $\beta$, so we can express all such polynomials as polynomial combinations of these. In particular, we find that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta = (-b)^2 - 2 c = b^2 - 2c$. So, for a monic polynomial, the ratio is
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{b^2 - 2c}{c} = \frac{b^2}{c} - 2$.
Evaluating for $b = -\frac{7}{2}$, $c = 6$ gives the result for your polynomial.
