Question

If 'm' and 'n' are the roots of equation $x^2-3x+4=0$ form the equation whose roots are $m^2$ and $n^2$.

Answer 1

Consider the equation $x^2-3x+4=0$. Here a = 1, b = - 3, c = 4
(i) Sum of the roots $=m+n=\frac{-b}{a}=\frac{-(-3)}{1}$
$\therefore m+n=3$
(ii) Product of the roots $mn=\frac{c}{a}=\frac{4}{1}$
$\therefore mn=4$
If the roots are $m^2$ and $n^2$
Sum of the roots $m^2+n^2=(m+n{)}^2-2mn=(3{)}^2-2\left(4\right)=9-8$
$\therefore m^2+n^2=1$
Product of the roots $m^2{n}^2=(mn{)}^2=4^2\therefore m^2{n}^2=16$
$x^2-(m^2+n^2)x+m^2{n}^2=0\therefore x^2-\left(1\right)x+\left(16\right)=0$
$x^2-x+16=0$