Question
Solve the following quadratic equations by factorization:
$9 x^2-3 x-2=0$
$9 x^2-3 x-2=0$
✓
Answer
We have
$9 x^2-3 x-2=0$
$\Rightarrow 9 x^2-6 x+3 x-2=0$
$\Rightarrow 3 x(3 x-2)+1(3 x-2)=0$
$\Rightarrow(3 x-2)(3 x+1)=0$
$\Rightarrow \text { either } 3 x-2=0 \text { or } 3 x+1=0$
$\Rightarrow 3 x=2 \text { or } 3 x=-1$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=-\frac{1}{3}$
Thus, $\text{x}=\frac{2}{3}$ and $\text{x}=-\frac{1}{3}$ are two roots of the equation $9 x^2-3 x-2=0$
$9 x^2-3 x-2=0$
$\Rightarrow 9 x^2-6 x+3 x-2=0$
$\Rightarrow 3 x(3 x-2)+1(3 x-2)=0$
$\Rightarrow(3 x-2)(3 x+1)=0$
$\Rightarrow \text { either } 3 x-2=0 \text { or } 3 x+1=0$
$\Rightarrow 3 x=2 \text { or } 3 x=-1$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=-\frac{1}{3}$
Thus, $\text{x}=\frac{2}{3}$ and $\text{x}=-\frac{1}{3}$ are two roots of the equation $9 x^2-3 x-2=0$
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