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

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Solution

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

Here, the given quadratic equation $x^{2} - 6 x + 2 = 0$ is in the form $a x^{2} + b x + c = 0$ 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:

$m n = \frac{c}{a} = \frac{2}{1} = 2$

Now, we find $m - n$ as follows:

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

Therefore,

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

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

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