Solve the following quadratic equations by factorization:

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

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Solution

$\frac{4}{x} - 3 = \frac{5}{2 x + 3} \Rightarrow \frac{4 - 3 x}{x} = \frac{5}{2 x + 3} \Rightarrow (4 - 3 x) (2 x + 3) = 5 x \Rightarrow 8 x + 12 - 6 x^{2} - 9 x = 5 x \Rightarrow - 6 x^{2} - 6 x + 12 = 0 \Rightarrow x^{2} + x - 2 = 0 \Rightarrow x^{2} + 2 x - x - 2 = 0 \Rightarrow x (x + 2) - 1 (x + 2) = 0 \Rightarrow (x - 1) (x + 2) = 0 \Rightarrow x - 1 = 0 or x + 2 = 0 \Rightarrow x = 1 or x = - 2$

Hence, the factors are 1 and −2.

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Similar questions

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

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

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