Question

Solve the following quadratic equation by completing the square: $x^2+(\sqrt{3}+1)x+\sqrt{3}=0$

• A. $\{\sqrt{3},1\}$
• B. $\{\sqrt{3},2\}$
• C. $\{\sqrt{3},3\}$
• D. None of these

Answer 1

The correct option is A $\{\sqrt{3},1\}$

Given equation is $x^2-(\sqrt{3}+1)x+\sqrt{3}=0$

$\Rightarrow$ $x^2-(\sqrt{3}+1)x=-\sqrt{3}$

Now, adding ${\left(\frac{\sqrt{3}+1}{2}\right)}^2$ on both sides,

$\Rightarrow$ $x^2-2\times x\times \left(\frac{\sqrt{3}+1}{2}\right)+{\left(\frac{\sqrt{3}+1}{2}\right)}^2=-\sqrt{3}+{\left(\frac{\sqrt{3}+1}{2}\right)}^2$

$\Rightarrow$ ${[x-\frac{(\sqrt{3}+1)}{2}]}^2=[\frac{(\sqrt{3}+1{)}^2}{4}-\sqrt{3}]$

$\Rightarrow$ ${[x-\frac{(\sqrt{3}+1)}{2}]}^2=\left[\frac{3+2\sqrt{3}+1-4\sqrt{3}}{4}\right]$

$\Rightarrow$ ${[x-\frac{(\sqrt{3}+1)}{2}]}^2=\left[\frac{3-2\sqrt{3}+1}{4}\right]$

$\Rightarrow$ ${[x-\frac{(\sqrt{3}+1)}{2}]}^2={\left[\frac{\sqrt{3}-1}{2}\right]}^2$

Taking square root on both sides,

$\Rightarrow$ $[x-\frac{(\sqrt{3}+1)}{2}]=\pm \frac{\sqrt{3}-1}{2}$

$\Rightarrow$ $[x-\frac{(\sqrt{3}+1)}{2}]=\frac{\sqrt{3}-1}{2}$

and $[x-\frac{(\sqrt{3}+1)}{2}]=-\frac{\sqrt{3}-1}{2}$

$\therefore$ $x=\frac{2\sqrt{3}}{2}$ and $x=\frac{2}{2}$

$\therefore$ $x=\sqrt{3}$ and $x=1$