Question
$(\text{x}+2)(\text{x}+3)+(\text{x}-3)(\text{x}-2)-2\text{x}(\text{x}+1)=0$
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Answer
$(\text{x}+2)(\text{x}+3)+(\text{x}-3)(\text{x}-2)-2\text{x}(\text{x}+1)=0$
$\Rightarrow[\text{x}^2+(2+3)\text{x}+2\times3]+[\text{x}^2+(-3-2)\text{x}\\+(-3)(-2)]-2\text{x}^2-2\text{x}=0$
$\Rightarrow\text{x}^2+5\text{x}+6+\text{x}^2-5\text{x}+6-2\text{x}^2-2\text{x}=0$
$\Rightarrow\text{x}^2+\text{x}^2-2\text{x}^2+5\text{x}-5\text{x}-2\text{x}+6+6=0$
$\Rightarrow-2\text{x}+12=0$
Subtracting 12 from both sides,
$-2\text{x}+12-12=0-12$
$\Rightarrow-2\text{x}=-12$
Dividing by - 2,
$\text{x}=6$
Verification:
$\text{L.H.S}=(\text{x}+2)(\text{x}+3)+(\text{x}-3)(\text{x}-2)-2\text{x}(\text{x}+1)$
$=(6+2)(6+3)+(6-3)(6-2)-2\times6\ (6+1)$
$=8\times9+3\times4-12 \times7$
$=84-84$
$=0$
$=\text{R.H.S}$
$\Rightarrow[\text{x}^2+(2+3)\text{x}+2\times3]+[\text{x}^2+(-3-2)\text{x}\\+(-3)(-2)]-2\text{x}^2-2\text{x}=0$
$\Rightarrow\text{x}^2+5\text{x}+6+\text{x}^2-5\text{x}+6-2\text{x}^2-2\text{x}=0$
$\Rightarrow\text{x}^2+\text{x}^2-2\text{x}^2+5\text{x}-5\text{x}-2\text{x}+6+6=0$
$\Rightarrow-2\text{x}+12=0$
Subtracting 12 from both sides,
$-2\text{x}+12-12=0-12$
$\Rightarrow-2\text{x}=-12$
Dividing by - 2,
$\text{x}=6$
Verification:
$\text{L.H.S}=(\text{x}+2)(\text{x}+3)+(\text{x}-3)(\text{x}-2)-2\text{x}(\text{x}+1)$
$=(6+2)(6+3)+(6-3)(6-2)-2\times6\ (6+1)$
$=8\times9+3\times4-12 \times7$
$=84-84$
$=0$
$=\text{R.H.S}$
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