Question

If 'm' and 'n' are the roots of equation ${}^2-3x+1=0$. Find the value of
(i) $m^{2}n+m{n}^2$ (ii) $\frac{1}{m}+\frac{1}{n}$.

Answer 1

Consider the equation $x^2-3x+1=0$
This is in the form $a{x}^2+bx+c=0$, where $a=1,b=-3,c=1$
(i) Sum of the roots $m+n=\frac{-b}{a}=\frac{-(-3)}{1}=3\therefore m+n=3$
(ii) Product of the roots $mn=\frac{c}{a}=\frac{1}{1}\therefore mn=1$
Therefore,
(i) $m^{2}n+m{n}^2=mn(m+n)=1\times 3=3$
(ii) $\frac{1}{m}+\frac{1}{n}=\frac{n+m}{mn}=\frac{m+n}{mn}=\frac{3}{1}=3\therefore \frac{1}{m}+\frac{1}{n}=3$