Question
Solve the following quadratic equations by factorization:
$\text{x}^2-\big(\sqrt{2}+1\big)\text{x}+\sqrt{2}=0$
$\text{x}^2-\big(\sqrt{2}+1\big)\text{x}+\sqrt{2}=0$
✓
Answer
$\text{x}^2-\big(\sqrt{2}+1\big)\text{x}+\sqrt{2}=0$
$\Rightarrow(\text{x})^2-2\Big(\frac{\sqrt{2}+1}{2}\Big)\text{x}+\sqrt{2}=0$
$\Rightarrow(\text{x})^2-2\times\Big(\frac{\sqrt{2}+1}{2}\Big)\times\text{x}+\Big(\frac{\sqrt{2}+1}{2}\Big)^2-\frac{3-2\sqrt{2}}{4}=0$
$\begin{cases}\Big(\frac{\sqrt{2}+1}{2}\Big)^2-\frac{3-2\sqrt{2}}{4}\\=\frac{2+1+2\sqrt{2}}{4}-\frac{3-2\sqrt{2}}{4}=\sqrt{2}\end{cases}$
$\Rightarrow\Big(\text{x}-\frac{\sqrt{2}+1}{2}\Big)^2$
$=\frac{3-2\sqrt{2}}{4}$
$=\frac{2+1-2\sqrt{2}}{4}$
$=\Big(\pm\frac{\sqrt{2}-1}{2}\Big)^2$
$\therefore\text{x}-\frac{\sqrt{2}+1}{2}=\pm\frac{\sqrt{2}-1}{2}$
$\Rightarrow\text{x}=\frac{\sqrt{2}+1}{2}\pm\frac{\sqrt{2}-1}{2}$
$\therefore\text{x}=\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}$
$=\frac{2\sqrt{2}}{2}-\sqrt{2}$
and $\text{x}=\frac{\sqrt{2}+1}{2}-\frac{\sqrt{2}-1}{2}$
$=\frac{2}{2}=1$
Hence roots are $1,\sqrt{2}$
$\Rightarrow(\text{x})^2-2\Big(\frac{\sqrt{2}+1}{2}\Big)\text{x}+\sqrt{2}=0$
$\Rightarrow(\text{x})^2-2\times\Big(\frac{\sqrt{2}+1}{2}\Big)\times\text{x}+\Big(\frac{\sqrt{2}+1}{2}\Big)^2-\frac{3-2\sqrt{2}}{4}=0$
$\begin{cases}\Big(\frac{\sqrt{2}+1}{2}\Big)^2-\frac{3-2\sqrt{2}}{4}\\=\frac{2+1+2\sqrt{2}}{4}-\frac{3-2\sqrt{2}}{4}=\sqrt{2}\end{cases}$
$\Rightarrow\Big(\text{x}-\frac{\sqrt{2}+1}{2}\Big)^2$
$=\frac{3-2\sqrt{2}}{4}$
$=\frac{2+1-2\sqrt{2}}{4}$
$=\Big(\pm\frac{\sqrt{2}-1}{2}\Big)^2$
$\therefore\text{x}-\frac{\sqrt{2}+1}{2}=\pm\frac{\sqrt{2}-1}{2}$
$\Rightarrow\text{x}=\frac{\sqrt{2}+1}{2}\pm\frac{\sqrt{2}-1}{2}$
$\therefore\text{x}=\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}$
$=\frac{2\sqrt{2}}{2}-\sqrt{2}$
and $\text{x}=\frac{\sqrt{2}+1}{2}-\frac{\sqrt{2}-1}{2}$
$=\frac{2}{2}=1$
Hence roots are $1,\sqrt{2}$
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