Solve this equation \[8x - \dfrac{3}{{3x}} = 2\].

Hint: Here, we need to find the value of x, which is an unknown variable.The given equation is not a linear equation. But it is a quadratic equation, with one variable.Any term of an equation may be taken from one side to other with the change in its sign, this does not affect the equality of the statement and this process is called transposition. We will use the quadratic method formula too\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step by step answer:
Given the equation as below,
\[8x - \dfrac{3}{{3x}} = 2\]
Solving this above equation, we will follow as below:
\[ \Rightarrow \dfrac{{8x(3x) - 3}}{{3x}} = 2\]
Taking the LCM in LHS, we get,
\[ \Rightarrow \dfrac{{24{x^2} - 3}}{{3x}} = 2\]
Removing the brackets, we get,
\[ \Rightarrow 24{x^2} - 3 = 2(3x)\]
By using transposition, the denominator value is moved to RHS, we get,
\[ \Rightarrow 24{x^2} - 3 = 6x\]
Again by using transposition, the RHS value is shifted to LHS, we get,
\[ \Rightarrow 24{x^2} - 3 - 6x = 0\]

Rearranging the above equation, we get,
\[ \Rightarrow 24{x^2} - 6x - 3 = 0\]
Divide the equation by 3, to minimize the numbers, we get,
\[ \Rightarrow 8{x^2} - 2x - 1 = 0\] ------- (i)
Using factorization method, we get,
\[ \Rightarrow 8{x^2} - 4x + 2x - 1 = 0\]
Simplify this equation, we get,
\[ \Rightarrow 4x(2x - 1) + 1(2x - 1) = 0\]
\[ \Rightarrow (4x + 1)(2x - 1) = 0\]
\[ \Rightarrow 4x + 1 = 0\]and \[2x - 1 = 0\]
\[ \Rightarrow 4x = - 1\] and \[2x = 1\]
\[ \therefore x = \dfrac{{ - 1}}{4}\] and \[x = \dfrac{1}{2}\]

Let us check the answer by substituting the value of $x$ in the given equation.Let's see if the answer is correct or not. First we will calculate the LHS part.So,
\[LHS=8x - \dfrac{3}{{3x}}\]
Substituting the value of \[x = \dfrac{1}{2}\], we get,
\[LHS= \dfrac{{8(\dfrac{1}{2})[3(\dfrac{1}{2})] - 3}}{{3(\dfrac{1}{2})}}\]
Dividing the number 8/2 we get,
\[LHS= \dfrac{{4[3(\dfrac{1}{2})] - 3}}{{3(\dfrac{1}{2})}}\]
Removing the brackets, we get,
\[LHS= \dfrac{{4(\dfrac{3}{2}) - 3}}{{\dfrac{3}{2}}}\]
\[\Rightarrow LHS= \dfrac{{\dfrac{{4(3)}}{2} - 3}}{{\dfrac{3}{2}}}\]
Taking LCM \[2\]in both numerator and denominator, we get,
\[LHS=\dfrac{{\dfrac{{4(3) - 3(2)}}{2}}}{{\dfrac{3}{2}}}\]

Removing the brackets, we get,
\[LHS= \dfrac{{\dfrac{{12 - 6}}{2}}}{{\dfrac{3}{2}}}\]
Simplify the expression, we get,
\[LHS= \dfrac{{\dfrac{6}{2}}}{{\dfrac{3}{2}}}\]
Divide the numerator value with the denominator value, we get,
\[LHS= \dfrac{6}{2} \div \dfrac{3}{2}\]
Convert the divide sign into multiplication sign, we get,
\[LHS= \dfrac{6}{2} \times \dfrac{2}{3}\]
Simplify this expression, we get,
\[LHS= \dfrac{6}{3}\]
\[\Rightarrow LHS= 2\]

Next, we will calculate the RHS part.
RHS \[ = 2\]
Hence, LHS = RHS and so the value of $x$ is correct. Thus, it is verified. In the same way, we can substitute \[x = \dfrac{{ - 1}}{4}\], to check if the answer is correct or not. Also, we can solve this using quadratic method, from equation (i), we get,
\[8{x^2} - 2x - 1 = 0\]
\[\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Where, \[a = 8,b = - 2,c = - 1\]
Substituting the values, we will get two values of $x$ as above.

Hence, the values of $x$ are $\dfrac{{ - 1}}{4}$ and $\dfrac{1}{2}$.

Note:The standard form of a linear equation with one variable is of the form ax + b = 0, where x is a variable, and ‘a’ and ‘b’ are constants. An equation containing a single variable of degree 2 is called Quadratic equation definition, which in the form\[a{x^2} + bx + c = 0\], where x is a variable and a, b and c are constants. We can check if the answer is correct or not by substituting the value of x in the given equation. We can solve this equation using the quadratic formula also.