Question

Form the quadratic equation whose roots are $\frac{1}{m}+\frac{1}{n}$

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,

$\frac{1}{m}+\frac{1}{n}=\frac{n+m}{mn}=\frac{m+n}{mn}=\frac{6}{2}=3$

Hence, the value of $\frac{1}{m}+\frac{1}{n}=3$.