Question
Solve the following quadratic equations by factorization:
$\frac{2\text{x}}{\text{x}-4}+\frac{2\text{x}-5}{\text{x}-3}=\frac{25}{3}$
$\frac{2\text{x}}{\text{x}-4}+\frac{2\text{x}-5}{\text{x}-3}=\frac{25}{3}$
✓
Answer
$\frac{2\text{x}}{\text{x}-4}+\frac{2\text{x}-5}{\text{x}-3}=\frac{25}{3}$$\Rightarrow\frac{2\text{x}(\text{x}-3)+(2\text{x}-5)(\text{x}-4)}{(\text{x}-4)(\text{x}-3)}=\frac{25}{3}$
$\Rightarrow\frac{2\text{x}^2-6\text{x}+2\text{x}^2-8\text{x}-5\text{x}+20}{\text{x}^2-3\text{x}-4\text{x}+12}=\frac{25}{3}$
$\Rightarrow\frac{4\text{x}^2-19\text{x}+20}{\text{x}^2-7\text{x}+12}=\frac{25}{3}$
$\Rightarrow 25x^2 - 175x + 300 = 12x^2 - 57x + 60$
$\Rightarrow 25x^2 - 175x + 300 - 12x^2 + 57x - 60= 0$
$\Rightarrow 13x^2 - 118x + 240 = 0$
$\Rightarrow 13x^2 - 78x - 40x + 240 = 0$
$\begin{cases}\because40\times13=3120\\\therefore3120=-78\times(-40)\\-118=-78-40\end{cases}$
⇒ 13x(x - 6) - 40(x - 6) = 0
⇒ (x - 6)(13x - 40) = 0
Either x - 6 = 0, then x = 6
or 13x - 40 = 0, then 13x = 40
$\Rightarrow\text{x}=\frac{40}{13}$
$\therefore$ Roots are 6, $\frac{40}{13}$
$\Rightarrow\frac{2\text{x}^2-6\text{x}+2\text{x}^2-8\text{x}-5\text{x}+20}{\text{x}^2-3\text{x}-4\text{x}+12}=\frac{25}{3}$
$\Rightarrow\frac{4\text{x}^2-19\text{x}+20}{\text{x}^2-7\text{x}+12}=\frac{25}{3}$
$\Rightarrow 25x^2 - 175x + 300 = 12x^2 - 57x + 60$
$\Rightarrow 25x^2 - 175x + 300 - 12x^2 + 57x - 60= 0$
$\Rightarrow 13x^2 - 118x + 240 = 0$
$\Rightarrow 13x^2 - 78x - 40x + 240 = 0$
$\begin{cases}\because40\times13=3120\\\therefore3120=-78\times(-40)\\-118=-78-40\end{cases}$
⇒ 13x(x - 6) - 40(x - 6) = 0
⇒ (x - 6)(13x - 40) = 0
Either x - 6 = 0, then x = 6
or 13x - 40 = 0, then 13x = 40
$\Rightarrow\text{x}=\frac{40}{13}$
$\therefore$ Roots are 6, $\frac{40}{13}$
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