Question

If '$m$' and '$n$' are the roots of the equation $x^2-6x+2=0$ find the value of $\frac{1}{n}-\frac{1}{m}$

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, we find $m-n$ as follows:

$(m-n{)}^2=(m+n{)}^2-4mn\Rightarrow (m-n{)}^2=6^2-(4\times 2)\Rightarrow (m-n{)}^2=36-8\Rightarrow (m-n{)}^2=28\Rightarrow m-n=\sqrt{28}\Rightarrow m-n=2\sqrt{7}$

Therefore,

$\frac{1}{n}-\frac{1}{m}=\frac{m-n}{nm}=\frac{2\sqrt{7}}{2}=\sqrt{7}$

Hence, the value of $\frac{1}{n}-\frac{1}{m}=\sqrt{7}$.