Question

Solve each of the following equations by using the method of completing the square:

$x^2-\left(\sqrt{2}+1\right)x+\sqrt{2}=0$

Answer 1

$$\begin{gathered} x^2-\left(\sqrt{2}+1\right)x+\sqrt{2}=0 \\ \Rightarrow x^2-\left(\sqrt{2}+1\right)x=-\sqrt{2} \\ \Rightarrow x^2-2\times x\times \left(\frac{\sqrt{2}+1}{2}\right)+{\left(\frac{\sqrt{2}+1}{2}\right)}^2=-\sqrt{2}+{\left(\frac{\sqrt{2}+1}{2}\right)}^2\left[\mathrm{Adding}{\left(\frac{\sqrt{2}+1}{2}\right)}^2\mathrm{on}\mathrm{both}\mathrm{sides}\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left[x-\left(\frac{\sqrt{2}+1}{2}\right)\right]}^2=\frac{-4\sqrt{2}+2+1+2\sqrt{2}}{4}=\frac{2-2\sqrt{2}+1}{4}={\left(\frac{\sqrt{2}-1}{2}\right)}^2 \\ \Rightarrow x-\left(\frac{\sqrt{2}+1}{2}\right)=\pm \left(\frac{\sqrt{2}-1}{2}\right)\left(\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow x-\left(\frac{\sqrt{2}+1}{2}\right)=\left(\frac{\sqrt{2}-1}{2}\right)\mathrm{or}x-\left(\frac{\sqrt{2}+1}{2}\right)=-\left(\frac{\sqrt{2}-1}{2}\right) \\ \Rightarrow x=\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\mathrm{or}x=\frac{\sqrt{2}+1}{2}-\frac{\sqrt{2}-1}{2} \end{gathered}$$
$\Rightarrow x=\frac{2\sqrt{2}}{2}=\sqrt{2}\mathrm{or}x=\frac{2}{2}=1$
Hence, $\sqrt{2}$ and 1 are the roots of the given equation.