Question
Solve $(x-10)\left(\frac{1200}{x}+2\right)=1260$ and $x<0$.
✓
Answer
Given quadratic equation is $(x-10)\left(\frac{1200}{x}+2\right)=1260$
$\Rightarrow(x-10)\left(\frac{1200}{x}+2\right)=1260$
$ \Rightarrow(x-10)(1200+2 x)=1260 x $
$ \Rightarrow 1200 x+2 x^2-12000-20 x+1260 x $
$ \Rightarrow 2 x^2-12000-8 x=0 $
$ \Rightarrow x^2-40 x-6000=0 $
$ \Rightarrow x^2-100 x+60 x-6000=0 $
$ \Rightarrow(x=100)(x+60)=0 $
$ \Rightarrow x =100 \text { or } x =-60$
But as $x < 0,$ so x can't be positive.
Hence, $x = -60$
$\Rightarrow(x-10)\left(\frac{1200}{x}+2\right)=1260$
$ \Rightarrow(x-10)(1200+2 x)=1260 x $
$ \Rightarrow 1200 x+2 x^2-12000-20 x+1260 x $
$ \Rightarrow 2 x^2-12000-8 x=0 $
$ \Rightarrow x^2-40 x-6000=0 $
$ \Rightarrow x^2-100 x+60 x-6000=0 $
$ \Rightarrow(x=100)(x+60)=0 $
$ \Rightarrow x =100 \text { or } x =-60$
But as $x < 0,$ so x can't be positive.
Hence, $x = -60$
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