Question

How do you solve rational equations $ \dfrac{2}{{3x + 1}} = \dfrac{1}{x} - \dfrac{{6x}}{{3x + 1}} $

Answer 1

Hint: In order to solve the above rational equation , simplify the equation by multiplying both sides of the equation with $ \left( x \right)\left( {3x + 1} \right) $ . Now rearrange the obtained quadratic equation in the form of a standard quadratic equation and then use the method of splitting the middle term to form the factors. Equate every factor with zero one by one to obtain the required solution.

Complete step-by-step answer:
We are given a rational equation in variable $ x $
 $ \dfrac{2}{{3x + 1}} = \dfrac{1}{x} - \dfrac{{6x}}{{3x + 1}} $
Simplifying the above equation by multiplying both sides of the equation with $ \left( x \right)\left( {3x + 1} \right) $ , we get
 $ \left( x \right)\left( {3x + 1} \right)\left( {\dfrac{2}{{3x + 1}}} \right) = \left( x \right)\left( {3x + 1} \right)\left( {\dfrac{1}{x} - \dfrac{{6x}}{{3x + 1}}} \right) $
For the right-hand side , use the distributive law of multiplication as $ A\left( {B - C} \right) = AB - AC $
 $ 2x = \left( x \right)\left( {3x + 1} \right)\left( {\dfrac{1}{x}} \right) - \left( x \right)\left( {3x + 1} \right)\left( {\dfrac{{6x}}{{3x + 1}}} \right) $
Simplifying it further , we have
 $ 2x = 3x + 1 - 6{x^2} $
Now rearrange the above quadratic equation in the form of standard quadratic equation as $ a{x^2} + bx + c $
 $ 6{x^2} + 2x - 3x - 1 = 0 $
Now taking common from the first 2 terms and last 2 terms , we get
 $ 2x\left( {3x + 1} \right) - 1\left( {3x + 1} \right) = 0 $
Finding the common binomial parenthesis, the equation becomes
 $ \left( {3x + 1} \right)\left( {2x - 1} \right) = 0 $
Now equating every factor with zero for the solution of $ x $
 $
  \left( {3x + 1} \right)\left( {2x - 1} \right) = 0 \\
   \Rightarrow 2x - 1 = 0 \\
  x = \dfrac{1}{2} \\
   \Rightarrow 3x + 1 = 0 \\
  x = \dfrac{{ - 1}}{3} \;
  $
 $ \therefore $ $ x = \dfrac{1}{2}, - \dfrac{1}{3} $
Therefore, the solution of the given rational equation is $ x = \dfrac{1}{2}, - \dfrac{1}{3} $ .
So, the correct answer is “$ x = \dfrac{1}{2}, - \dfrac{1}{3} $”.

Note: Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $ a{x^2} + bx + c $ where $ x $ is the unknown variable and a,b,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $ D = {b^2} - 4ac $
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions