Solve for x: $\dfrac{2}{{3{x^2}}} - \dfrac{1}{{3x}} - 1 = 0$

Hint: In this question, we are given an algebraic expression in which the unknown variables are in the denominator so we will first simplify the expression by taking the least common factor of the denominators of the three fractions ( LCM is the smallest number divisible by both the numbers}. After finding the common denominator for the three fractions, the numerators are multiplied with the quotients obtained on dividing the LCM by their denominator and then we have to perform the given arithmetic operation like addition, subtraction, multiplication and division in the numerator.

Complete step-by-step solution:
We have to solve $\dfrac{2}{{3{x^2}}} - \dfrac{1}{{3x}} - 1 = 0$
It can be rewritten as –
$
 \Rightarrow \dfrac{2}{{3{x^2}}} - \dfrac{1}{{3x}} - \dfrac{1}{1} = 0 \\
   \Rightarrow \dfrac{{2 - x - 3{x^2}}}{{3{x^2}}} = 0 \\
 $
Taking the negative side common and then taking -1 and $3{x^2}$ to the other side, we get –
$3{x^2} + x - 2 = 0$
The obtained equation is a quadratic equation and can be solved by factorization as follows –
$
 \Rightarrow 3{x^2} + 3x - 2x - 2 = 0 \\
   \Rightarrow 3x(x + 1) - 2(x + 1) = 0 \\
   \Rightarrow (3x - 2)(x + 1) = 0 \\
   \Rightarrow x = \dfrac{2}{3},\,x = - 1 \\
 $
Hence, when $\dfrac{2}{{3{x^2}}} - \dfrac{1}{{3x}} - 1 = 0$ , we get $x = \dfrac{2}{3}$or $x = - 1$ .

Note: The highest exponent is 2 so the given equation is a polynomial equation of degree 2, hence it is a quadratic equation and has exactly two solutions. We have to solve this equation, that is, we have to find its solutions. $a{x^2} + bx + c = 0$ is the standard form of a quadratic equation. To find the factors of the given equation, we compare the given equation and the standard equation and get the values of a, b and c. Then we will try to write b as a sum of two numbers such that their product is equal to the product of a and c, that is, ${b_1} \times {b_2} = a \times c$ , this method is known as factorization. We find the value of ${b_1}$ and ${b_2}$ by hit and trial.