Question

Solve by factorization method: $\left( {\dfrac{4}{x}} \right) - 3 = \dfrac{5}{{(2x + 3)}}$ ,$x \ne 0$ ,$ - \dfrac{3}{2}$

Answer 1

Hint: Generally, three methods are used for solving quadratic equations. They are the factorization method, completing the square method, and using the quadratic formula. Here, we are asked to solve the given quadratic equation by using the factorization method. Now, we shall learn about the factorization method so that we are able to solve the problem.
             To solve a quadratic equation by the factorization method, first, try to put all the terms on one side of the equal sign, and the other side of the equal sign must be zero. Then make the opposite side to contain the factors of the given polynomial and then set each factor equal to zero. Finally, we will get our required solution.

Complete step by step answer:
We are given an equation$\left( {\dfrac{4}{x}} \right) - 3 = \dfrac{5}{{(2x + 3)}}$
We are asked to solve it by using the factorization method.
$\left( {\dfrac{4}{x}} \right) - 3 = \dfrac{5}{{(2x + 3)}}$
We shall take LCM on the left-hand side
$ \Leftarrow \dfrac{{4 - 3x}}{x} = \dfrac{5}{{(2x + 3)}}$
Now, we need to cross multiply the above equation.
$ \Leftarrow \left( {4 - 3x} \right)(2x + 3) = 5x$
We shall multiply the terms on the left-hand side.
$ \Rightarrow 8x + 12 - 6{x^2} - 9x = 5x$
$ \Rightarrow - 6{x^2} - x + 12 = 5x$
$ \Rightarrow - 6{x^2} - x + 12 - 5x = 0$
$ \Rightarrow - 6{x^2} - 6x + 12 = 0$
We shall multiply $ - $ on both sides.
$ \Rightarrow 6{x^2} + 6x - 12 = 0$
Now, we have to multiply $\dfrac{1}{6}$ into both sides.
$ \Rightarrow \dfrac{1}{6}6{x^2} + \dfrac{1}{6}6x - \dfrac{1}{6}12 = 0 \times \dfrac{1}{6}$
$ \Rightarrow {x^2} + x - 2 = 0$
We shall split the term$x$ as$2x - x$ .
$ \Rightarrow {x^2} + 2x - x - 2 = 0$
Now, we need to pick the common terms.
$ \Rightarrow x\left( {x + 2} \right) - 1\left( {x + 2} \right) = 0$
$ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right) = 0$
We get$\left( {x - 1} \right) = 0$ or$\left( {x + 2} \right) = 0$
Hence, $x = 1$ or$x = - 2$ is the required solution.

Note: The factorization method, completing the square method, and using the quadratic formula are the three methods to solve a quadratic equation. If we are not given any details about the method in the question, we can apply any method to solve the given quadratic equation. Among the three methods, the factorization method is usually preferred.