Question

Solve for x by completing the square method:
$x^2-(\sqrt{3}+1)x+\sqrt{3}=0$

• A. 1 and $\sqrt{3}$
• B. 2 and $\sqrt{3}$
• C. 1 and 2
• D. 3 and 4

Answer 1

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

We have,
$x^2-(\sqrt{3}+1)x+\sqrt{3}=0$

$\Rightarrow x^2-\frac{2(\sqrt{3}+1)}{2}.x+\sqrt{3}=0$

$\Rightarrow x^2-2.x.\frac{(\sqrt{3}+1)}{2}+\sqrt{3}=0$

$\Rightarrow x^2-2.x.\left(\frac{\sqrt{3}+1}{2}\right)+(\frac{\sqrt{3}+1}{2}{)}^2-(\frac{\sqrt{3}+1}{2}{)}^2+\sqrt{3}=0$

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

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

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

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

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

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

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

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

$x=1andx=\sqrt{3}$

So, a is the correct option.