Question
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}+4}-\frac{1}{\text{x}-7}=\frac{11}{30},$ $\text{x}\neq4,7$
$\frac{1}{\text{x}+4}-\frac{1}{\text{x}-7}=\frac{11}{30},$ $\text{x}\neq4,7$
✓
Answer
$\frac{1}{\text{x}+4}-\frac{1}{\text{x}-7}=\frac{11}{30},$ $\text{x}\neq4,7$
$\frac{\text{x}-7-\text{x}-4}{(\text{x}+4)(\text{x}-7)}=\frac{11}{30}$
$\frac{-11}{(\text{x}+4)(\text{x}-7)}=\frac{11}{30}$
$\frac{-1}{(\text{x}+4)(\text{x}-7)}=\frac{1}{30}$ (dividing both side by 11)
$ (x+4)(x-7)=-30$
$ x^2+4 x-7 x-28+30=0 $
$ \Rightarrow x^2-3 x+2=0$
$\begin{Bmatrix}\because2=(-2)\times(-1)\\-3=-2-1\end{Bmatrix}$
$\Rightarrow x^2-x-2 x+2=0 $
$ \Rightarrow x(x-1)-2(x-1)=0 $
$\Rightarrow(x-1)(x-2)=0$
Either $x - 1 = 0$, then $x = 1$
Or $x - 2 = 0,$ then $x = 2$
$\therefore x = 1, 2$
$\frac{\text{x}-7-\text{x}-4}{(\text{x}+4)(\text{x}-7)}=\frac{11}{30}$
$\frac{-11}{(\text{x}+4)(\text{x}-7)}=\frac{11}{30}$
$\frac{-1}{(\text{x}+4)(\text{x}-7)}=\frac{1}{30}$ (dividing both side by 11)
$ (x+4)(x-7)=-30$
$ x^2+4 x-7 x-28+30=0 $
$ \Rightarrow x^2-3 x+2=0$
$\begin{Bmatrix}\because2=(-2)\times(-1)\\-3=-2-1\end{Bmatrix}$
$\Rightarrow x^2-x-2 x+2=0 $
$ \Rightarrow x(x-1)-2(x-1)=0 $
$\Rightarrow(x-1)(x-2)=0$
Either $x - 1 = 0$, then $x = 1$
Or $x - 2 = 0,$ then $x = 2$
$\therefore x = 1, 2$
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