Question
Solve $: (a+b)^2 x^2-(a+b) x-6=0 ; a+b \neq 0$
✓
Answer
$(a+b)^2 x^2-(a+b) x-6=0 ; a+b \neq 0$
$\Rightarrow(a+b)^2 x^2-3(a+b) x+2(a+b) x-6=0$
$ \Rightarrow(a+b) x[(a+b) x-3]+2[(a+b) x-3]=0$
$\Rightarrow[(a+b) x-3][(a+b) x+2]=0 $
$ \Rightarrow (a+b) x-3=0 \text { or }(a+b) x+2=0$
$\Rightarrow x=\frac{3}{a+b}$ or $x=\frac{-2}{a+b}$
$\Rightarrow(a+b)^2 x^2-3(a+b) x+2(a+b) x-6=0$
$ \Rightarrow(a+b) x[(a+b) x-3]+2[(a+b) x-3]=0$
$\Rightarrow[(a+b) x-3][(a+b) x+2]=0 $
$ \Rightarrow (a+b) x-3=0 \text { or }(a+b) x+2=0$
$\Rightarrow x=\frac{3}{a+b}$ or $x=\frac{-2}{a+b}$
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