2017/024)If m is a root of $A(x)=x^2−nx+10=0$, and n is a root of $B(x)=x−mx+20=0$, what is the value of $(\frac{n}{m} + \frac{m}{n} )^2$

m is root of  $A(x)=x^2−nx+10=0$
so $m^2-mn+10= 0$

n is root of $B(x)=x−mx+20=0$ hence  $n^2-mn+20 = 0$

so we have $m(m-n) = - 10$ and $n(m-n) = 20$ or n = -2m

so $\frac{n}{m} = -2$ and $\frac{m}{n} = -\frac{1}{2}$ or $(\frac{n}{m} + \frac{m}{n})^2 = \frac{25}{4}$