Solve the given quadratic equation by factorization method
$\frac{4}{x} - 3 = \frac{5}{2 x + 3} x \neq 0, \frac{- 3}{2}$

Open in App

Solution

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

$\Rightarrow \frac{4 - 3 x}{x} = \frac{5}{2 x + 3} \Rightarrow (4 - 3 x) (2 x + 3) = 5 x \Rightarrow 12 - x - 6 x^{2} = 5 x$

$\Rightarrow 6 x^{2} + 6 x - 12 = 0 \Rightarrow x^{2} + x - 2 = 0 \Rightarrow x^{2} + 2 x - x - 2 = 0 \Rightarrow x (x + 2) - (x + 2) = 0$

$\Rightarrow (x + 2) (x - 1) = 0 \Rightarrow x + 2 = 0 o r x - 1 = 0$

$\Rightarrow x = - 2 o r x = 1$

Suggest Corrections

Similar questions

\frac{2 x}{x - 3} + \frac{1}{2 x + 3} + \frac{3 x + 9}{(x - 3) (2 x + 3)} = 0

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

\sqrt{5} x^{2} + 2 x - 3 \sqrt{5} = 0

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

\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}

Join BYJU'S Learning Program