Rewrite the following as a quadratic equation in x and then solve for x:
$\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, - \frac{3}{2} .$

Open in App

Solution

Step 1:
Express the given equation in quadratic form.
$\begin{array}{rcl} \frac{4}{x} - 3 & = & \frac{5}{2 x + 3} \\ \Rightarrow & \frac{4 - 3 x}{x} = \frac{5}{2 x + 3} \\ \Rightarrow & (4 - 3 x) (2 x + 3) = 5 x \\ \Rightarrow & 8 x + 12 - 6 x^{2} - 9 x = 5 x \\ \Rightarrow & 6 x^{2} + 6 x - 12 = 0 \\ \Rightarrow & x^{2} + x - 2 = 0 \end{array}$
Step 2:
Solve the quadratic equation:
Use factorization method:
$\begin{array}{rcl} \Rightarrow & x^{2} + 2 x - x - 2 = 0 \\ \Rightarrow & x (x + 2) - 1 (x + 2) = 0 \\ \Rightarrow & (x + 2) (x - 1) = 0 \\ \Rightarrow & x + 2 = 0 o r x - 1 = 0 \\ \Rightarrow & x = - 2 o r x = 1 \end{array}$
So, the required value is $x = 1 and x = - 2$ .
Final answer:
The required value is $x = 1 and x = - 2$ .

Suggest Corrections

Similar questions

Solve each of the following quadratic equations:
$\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, \frac{- 3}{2}$

Solve the following quadratic equation by factorization.
$\frac{2}{x + 1} + \frac{3}{2 (x - 2)} = \frac{23}{5 x}; x \neq 0, - 1, 2$

Solve the following quadratic equation by factorization.
$\frac{x + 3}{x + 2} = \frac{3 x - 7}{2 x - 3}, x \neq - 2, \frac{3}{2}$

Solve the following quadratic equation by factorization.
(2x+3) (3x-7) = 0

\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, \frac{- 3}{2}

Join BYJU'S Learning Program