Question
Solve the following quadratic equations by factorization:
$\frac{4}{\text{x}}-3=\frac{5}{2\text{x}+3},$ $\text{x}\neq-0,-\frac{3}{2}$
$\frac{4}{\text{x}}-3=\frac{5}{2\text{x}+3},$ $\text{x}\neq-0,-\frac{3}{2}$
✓
Answer
$\frac{4}{\text{x}}-3=\frac{5}{2\text{x}+3}$
$\Rightarrow\frac{4-3\text{x}}{\text{x}}=\frac{5}{2\text{x}+3}$
$\Rightarrow (4 - 3x)(2x + 3) = 5x$
$\Rightarrow 8x + 12 - 6x^2 - 9x = 5x$
$\Rightarrow -6x^2 - 6x + 12 = 0$
$\Rightarrow x^2 + x - 2 = 0$
$\Rightarrow x^2 + 2x - x - 2 = 0$
$\Rightarrow x(x + 2) - 1(x + 2) = 0$
$\Rightarrow (x - 1)(x + 2) = 0$
$\Rightarrow x - 1 = 0$ or $x + 2 = 0$
$\Rightarrow x = 1$ or $x = -2$
Hence, the factors are $1$ and $-2$
$\Rightarrow\frac{4-3\text{x}}{\text{x}}=\frac{5}{2\text{x}+3}$
$\Rightarrow (4 - 3x)(2x + 3) = 5x$
$\Rightarrow 8x + 12 - 6x^2 - 9x = 5x$
$\Rightarrow -6x^2 - 6x + 12 = 0$
$\Rightarrow x^2 + x - 2 = 0$
$\Rightarrow x^2 + 2x - x - 2 = 0$
$\Rightarrow x(x + 2) - 1(x + 2) = 0$
$\Rightarrow (x - 1)(x + 2) = 0$
$\Rightarrow x - 1 = 0$ or $x + 2 = 0$
$\Rightarrow x = 1$ or $x = -2$
Hence, the factors are $1$ and $-2$
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