Question

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

• A. $x=\sqrt{1},3$
• B. $x=\sqrt{5},1$
• C. $x=\sqrt{1}0,2$
• D. $x=\sqrt{3},1$

Answer 1

The correct option is D $x=\sqrt{3},1$

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

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

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

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

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

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

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

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

$\Rightarrow x=\sqrt{3},1$

Hence, the roots are $\sqrt{3}$ and $1$.