Question
Solve: $x^2+5 x-\left(a^2+a-6\right)=0$
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Answer
$\text { Given, } x^2+5 x-\left(a^2+a-6\right)=0$
$\text { splitting } a^2+a-6$
$\Rightarrow x^2+5 x-\left(a^2+3 a-2 a-6\right)=0$
$\Rightarrow x^2+5 x-[a(a+3)-2(a+3)]=0$
$\Rightarrow x^2+5 x-(a+3)(a-2)=0$
Now splitting the middle term
$\Rightarrow x^2+(a+3) x-(a-2) x-(a+3)(a-2)=0$
$\Rightarrow x[x+(a+3)]-(a-2)[x+(a+3)]=0$
$\Rightarrow[x+(a+3)][x-(a-2)]=0$
$\Rightarrow x+(a+3)=0 \text { or } x-(a-2)=0$
Therefore, $x=-(a+3)$ or $(a-2)$
$\text { splitting } a^2+a-6$
$\Rightarrow x^2+5 x-\left(a^2+3 a-2 a-6\right)=0$
$\Rightarrow x^2+5 x-[a(a+3)-2(a+3)]=0$
$\Rightarrow x^2+5 x-(a+3)(a-2)=0$
Now splitting the middle term
$\Rightarrow x^2+(a+3) x-(a-2) x-(a+3)(a-2)=0$
$\Rightarrow x[x+(a+3)]-(a-2)[x+(a+3)]=0$
$\Rightarrow[x+(a+3)][x-(a-2)]=0$
$\Rightarrow x+(a+3)=0 \text { or } x-(a-2)=0$
Therefore, $x=-(a+3)$ or $(a-2)$
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