Solve $\dfrac{2}{{{x}^{2}}}-\dfrac{5}{x}+2=0$ by factorization method. Find the roots of (PS) $\dfrac{1}{x}-\dfrac{1}{x-2}=3x\ne 0,2$ .

Hint: Take the least common multiple in the equation and send the denominator to the right hand side. By this we get a quadratic equation in x. Try to divide the x term into a sum of 2 terms which can make the quadratic into a product of 2 single degree expressions. By this find the value of x. For the second equation also take the least common multiple and send the denominator to the right hand side. Repeat the same process, to the quadratic you get after least common multiple of get the values of x. These 4 values of x are the required result.

Complete step-by-step solution -
Factorization: For factoring a quadratic follow the steps:
Allot ${{x}^{2}}$ coefficient as “a”, x co-efficient as “b”, constant as “c”.
Find the product of the 2 numbers a, c. Let it be k.
Find 2 numbers such that their product is k, sum is b.
Rewrite the term $bx$ in terms of those 2 numbers.
Now, factor the first two terms.
Now, factor the last two terms.
If we do it correctly our 2 new terms will have one more common factor.
Now take that common to get quadratic as a product of 2 terms, from which you get the roots.
First expression we have in the question is given by:
$\dfrac{2}{{{x}^{2}}}-\dfrac{5}{x}+2=0$
By taking least common multiple, we get a quadratic given by:
$\Rightarrow 2-5x+2{{x}^{2}}=0$
Here, we get $a=2,b=-5,c=2$ . So, product ac is: 4.
So, 2 numbers whose product is 4 and sum is $-5$ are given by $-1,-4$ $\left( \because -1\times -4=4;-4+\left( -1 \right)=-5 \right)$
By this our quadratic equation turns into the form:
$2{{x}^{2}}-4x-x+2=0$
Now take 2x common from first two terms we get:
$2x\left( x-2 \right)-x+2=0$
By taking $\left( x-2 \right)$ common from whole expression, we get:
$\left( x-2 \right)\left( 2x-1 \right)=0$
By this we get 2 equations, which are written as:
$x-2=0$ $2x-1=0$
From these, we can write the roots as given below:
$x=2;x=\dfrac{1}{2}$
Second expression we have in the question is given by:
$\Rightarrow \dfrac{1}{x}-\dfrac{1}{\left( x-2 \right)}=3$
By multiplying $x\left( x-2 \right)$ on both sides of equation, we get:
$\Rightarrow \dfrac{x\left( x-2 \right)}{x}-\dfrac{x\left( x-2 \right)}{x-2}=3x\left( x-2 \right)$
By simplifying the above equation, we get an equation:
$\Rightarrow x-2-x=3{{x}^{2}}-6x$
By adding 2 on both sides of equation, we get
$\Rightarrow 3{{x}^{2}}-6x+2=0$
So, the 2 numbers whose product is 6, sum is $-6$ cannot be found naturally.
So, we use formula of quadratic equation roots of $a{{x}^{2}}+bx+c=0$ are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
By substituting a, b, c, values into this equation, we get:
$\Rightarrow x=\dfrac{6\pm \sqrt{36-12}}{6}$
By simplifying term inside the root, we get x value as:
$\Rightarrow x=\dfrac{6\pm \sqrt{12}}{6}=\dfrac{6\pm 2\sqrt{3}}{6}$
By taking 2 common and cancelling it in denominator we get:
$\Rightarrow x=\dfrac{3\pm \sqrt{3}}{3}$
So, the roots of the equation are $\dfrac{3+\sqrt{3}}{3},\dfrac{3-\sqrt{3}}{3}$ .
Therefore, $2,\dfrac{1}{2}$ are roots of first expression and $\dfrac{3+\sqrt{3}}{3},\dfrac{3-\sqrt{3}}{3}$ are roots of second expression.

Note: Alternate method instead of writing $-4x-x$ you can write $-x-4x$ take x common in first two terms and $-2$ common in last terms. Anyways, you will get the same result. Whenever you can’t divide any 2 integers for the condition of product ac and sum b you just use the formula of quadratic equation roots. It will always give the correct answer.