Question

Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
$x^2-(\sqrt{2}+1)x+\sqrt{2}=0$

• A. $exist,\sqrt{2}$, $1$
• B. $exist,-\sqrt{2}$, $1$
• C. $exist,-\sqrt{2}$, $-1$
• D. $\text{does not exist}$

Answer 1

Answer: (A)

$exist,\sqrt{2}$, $1$

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

Here, $a=1,b=-(\sqrt{2}+1),c=\sqrt{2}$

$\Rightarrow$ $x^2-(\sqrt{2}+1)x=-\sqrt{2}$

Now, ${\left(\frac{\sqrt{2}+1}{2}\right)}^2=\frac{2+2\sqrt{2}+1}{4}$

Adding $\frac{2+2\sqrt{2}+1}{4}$, on both sides,

$\Rightarrow$ $x^2-(\sqrt{2}+1)x+\frac{2+2\sqrt{2}+1}{4}=-\sqrt{2}+\frac{2+2\sqrt{2}+1}{4}$

$\Rightarrow$ ${[x-\left(\frac{\sqrt{2}+1}{2}\right)]}^2=\frac{2-2\sqrt{2}+1}{4}$

$\Rightarrow$ ${[x-\left(\frac{\sqrt{2}+1}{2}\right)]}^2={\left(\frac{\sqrt{2}-1}{2}\right)}^2$

$\Rightarrow$ $x-\left(\frac{\sqrt{2}+1}{2}\right)=\pm \left(\frac{\sqrt{2}-1}{2}\right)$

$\Rightarrow$ $x=\frac{\sqrt{2}-1+\sqrt{2}+1}{2}$ and $x=\frac{-\sqrt{2}+1+\sqrt{2}+1}{2}$

$\therefore$ $x=\sqrt{2}$ and $x=1$