Question
Solve the following quadratic equation by factorisation method:
$\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}$
$\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}$
✓
Answer
$\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4} $
$ \Rightarrow \frac{x^2+3 x-(x-2)(1-x)}{x(x-2)}=\frac{17}{4} $
$ \Rightarrow \frac{x^2+3 x-\left(x-x^2-2+2 x\right)}{x^2-2 x}=\frac{17}{4} $
$ \Rightarrow \frac{x^2+3 x-\left(-x^2+3 x-2\right)}{x^2-2 x}=\frac{17}{4} $
$ \Rightarrow \frac{x^2+3 x+x^2-3 x+2}{x^2-2 x}=\frac{17}{4} $
$ \Rightarrow \frac{2 x^2+2}{x^2-2 x}=\frac{17}{4}$
$\Rightarrow 17 x^2-34 x-8 x^2+8 $
$ \Rightarrow 9 x^2-34 x-8=0 $
$ \Rightarrow 9 x^2-36 x+2 x-8=0 $
$ \Rightarrow 9 x(x-4)+2(x-4)=0 $
$ \Rightarrow(x-4)(9 x+2)=0 $
$ \Rightarrow x-4=0 \text { or } 9 x+2=0 $
$ \Rightarrow x=4 \text { or } x=-\frac{2}{9} .$
$ \Rightarrow \frac{x^2+3 x-(x-2)(1-x)}{x(x-2)}=\frac{17}{4} $
$ \Rightarrow \frac{x^2+3 x-\left(x-x^2-2+2 x\right)}{x^2-2 x}=\frac{17}{4} $
$ \Rightarrow \frac{x^2+3 x-\left(-x^2+3 x-2\right)}{x^2-2 x}=\frac{17}{4} $
$ \Rightarrow \frac{x^2+3 x+x^2-3 x+2}{x^2-2 x}=\frac{17}{4} $
$ \Rightarrow \frac{2 x^2+2}{x^2-2 x}=\frac{17}{4}$
$\Rightarrow 17 x^2-34 x-8 x^2+8 $
$ \Rightarrow 9 x^2-34 x-8=0 $
$ \Rightarrow 9 x^2-36 x+2 x-8=0 $
$ \Rightarrow 9 x(x-4)+2(x-4)=0 $
$ \Rightarrow(x-4)(9 x+2)=0 $
$ \Rightarrow x-4=0 \text { or } 9 x+2=0 $
$ \Rightarrow x=4 \text { or } x=-\frac{2}{9} .$
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