Question
Solve the following quadratic equations by factorization:
$\frac{\text{x}+3}{\text{x}+2}=\frac{3\text{x}-7}{2\text{x}-3}$
$\frac{\text{x}+3}{\text{x}+2}=\frac{3\text{x}-7}{2\text{x}-3}$
✓
Answer
We have, $\frac{\text{x}+3}{\text{x}+2}=\frac{3\text{x}-7}{2\text{x}-3}$
$\Rightarrow (x + 3)(2x - 3) = (x + 2)(3x - 7)$
$\Rightarrow 2x^2 - 3x + 6x - 9 = 3x^2 - x - 14$
$\Rightarrow 2x^2 + 3x - 9 = 3x^2 - x - 14$
$\Rightarrow x^2 - 3x - x - 14 + 9 = 0$
$\Rightarrow x^2 - 4x - 5 = 0$
$[1x - 5 = -5 - 4 = -5 + 1]$
$\Rightarrow x^2 - 5x + x - 5 = 0$
$\Rightarrow x(x - 5) + 1(x - 5) = 0$
$\Rightarrow (x - 5)(x + 1) = 0$
$\Rightarrow x - 5 = 0 or x + 1 = 0$
$\Rightarrow x = 5 = 0 or x = -1$
$\therefore$ x = 5 and x = -1 are the two roots of the given quadratic equation.
$\Rightarrow (x + 3)(2x - 3) = (x + 2)(3x - 7)$
$\Rightarrow 2x^2 - 3x + 6x - 9 = 3x^2 - x - 14$
$\Rightarrow 2x^2 + 3x - 9 = 3x^2 - x - 14$
$\Rightarrow x^2 - 3x - x - 14 + 9 = 0$
$\Rightarrow x^2 - 4x - 5 = 0$
$[1x - 5 = -5 - 4 = -5 + 1]$
$\Rightarrow x^2 - 5x + x - 5 = 0$
$\Rightarrow x(x - 5) + 1(x - 5) = 0$
$\Rightarrow (x - 5)(x + 1) = 0$
$\Rightarrow x - 5 = 0 or x + 1 = 0$
$\Rightarrow x = 5 = 0 or x = -1$
$\therefore$ x = 5 and x = -1 are the two roots of the given quadratic equation.
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