Question

Solve the following quadratic equation by factorization method:
\[\dfrac{4}{x} - 3 = \dfrac{5}{{2x + 3}}\]

Answer 1

Hint: Here, we will first convert the given quadratic equation in standard form. Then we will factorize the quadratic equation. We will first split the middle term into two terms such that their product will be equal to the product of the first and last term. Then we will take common factors from the first two terms and the last two terms. We will then apply the zero product rule to find the required solution.

Complete step by step solution:
We will convert the given equation, \[\dfrac{4}{x} - 3 = \dfrac{5}{{2x + 3}}\], in standard form. Therefore, we get
\[ \Rightarrow \dfrac{{4 - 3x}}{x} = \dfrac{5}{{2x + 3}}\]
On cross multiplying the terms, we get
\[ \Rightarrow \left( {4 - 3x} \right)\left( {2x + 3} \right) = 5x\]
Multiplying the terms, we get
\[ \Rightarrow 8x - 6{x^2} + 12 - 9x = 5x\]
Adding and subtracting the like terms, we get
\[ \Rightarrow 6{x^2} + 6x - 12 = 0\]
Dividing both sides by 6, we get
\[ \Rightarrow {x^2} + x - 2 = 0\]
Now, the quadratic equation is in standard form.
We will now factorize the above equation by splitting the middle term into two terms.
Therefore,
\[ \Rightarrow {x^2} + 2x - x - 2 = 0\]
We can see that the product of the middle two terms and the product of the first and last term are equal to \[2{x^2}\].
Now, we will factorize the first two terms and last two terms separately.
\[ \Rightarrow x\left( {x + 2} \right) - \left( {x + 2} \right) = 0\]
We can see that both new terms have a common factor.
Therefore,
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right) = 0\]
Applying zero product property, we get
 \[\begin{array}{l} \Rightarrow x - 1 = 0\\ \Rightarrow x = 1\end{array}\]
or
\[\begin{array}{l} \Rightarrow x + 2 = 0\\ \Rightarrow x = - 2\end{array}\] .
Hence, the possible values of \[x\] are 1 and \[ - 2\].

Note: The equation we formed is a quadratic equation. An equation is said to be a quadratic equation if the highest degree of the variable is 2. The degree 2 suggests that there are 2 roots of the given quadratic equation. Roots of the equations are those values, which when substituted in the equation, then the equation becomes zero. Therefore we have to form the equations in such a way that we will get the value of the variable \[x\]. The value of \[x\]are the roots of the equation.
We have also applied zero product property. Zero product property states that if \[a \cdot b = 0\], then either \[a = 0\] or \[b = 0\].