Question
Solve the following quadratic equations by factorization:
$\text{x}^2-\Big(\sqrt{3}+1\Big)\text{x}+\sqrt{3}=0$
$\text{x}^2-\Big(\sqrt{3}+1\Big)\text{x}+\sqrt{3}=0$
✓
Answer
$\text{x}^2-\Big(\sqrt{3}+1\Big)\text{x}+\sqrt{3}=0$$\Rightarrow\text{x}^2-\sqrt{3}\text{x}-\text{x}+\sqrt{3}=0$
$\Rightarrow\text{x}\Big(\text{x}-\sqrt{3}\Big)-1\Big(\text{x}-\sqrt{3}\Big)=0$
$\Rightarrow\Big(\text{x}-\sqrt{3}\Big)(\text{x}-1)=0$
Either $\text{x}-\sqrt{3}=0,$ then $\text{x}=\sqrt{3}$
or $\text{x}-1=0,$ then $\text{x}=1$
$\therefore$ Roots are $\text{x}=\sqrt{3},1$
$\Rightarrow\text{x}\Big(\text{x}-\sqrt{3}\Big)-1\Big(\text{x}-\sqrt{3}\Big)=0$
$\Rightarrow\Big(\text{x}-\sqrt{3}\Big)(\text{x}-1)=0$
Either $\text{x}-\sqrt{3}=0,$ then $\text{x}=\sqrt{3}$
or $\text{x}-1=0,$ then $\text{x}=1$
$\therefore$ Roots are $\text{x}=\sqrt{3},1$
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