Question

If 'm' and 'n' are the roots of the equation $x^2-6x+2=0$ find the value of $m^3{n}^2+n^3{m}^2$

Answer 1

We know that if $m$ and $n$ are the roots of a quadratic equation $a{x}^2+bx+c=0$, the sum of the roots is $m+n=-\frac{b}{a}$ and the product of the roots is $mn=\frac{c}{a}$.

Here, the given quadratic equation $x^2-6x+2=0$ is in the form $a{x}^2+bx+c=0$ where $a=1,b=-6$ and $c=2$.

The sum of the roots is:

$m+n=-\frac{b}{a}=-\frac{(-6)}{1}=6$

The product of the roots is $\frac{c}{a}$ that is:

$mn=\frac{c}{a}=\frac{2}{1}=2$

Now,

$m^3{n}^2+n^3{m}^2=m^2{n}^2(m+n)=\left(mn{)}^2\right(m+n)=2^2\times 6=4\times 6=24$

Hence, the value of $m^3{n}^2+n^3{m}^2=24$.