Question

If $2-\sqrt{3}$ is a root of the quadratic equation $x^2+2(\sqrt{3}-1)x+3-2\sqrt{3}=0$ then the second root is

• A. $\sqrt{3}-2$
• B. $\sqrt{3}$
• C. $2+\sqrt{3}$
• D. $-\sqrt{3}$

Answer 1

Answer: (D)

$-\sqrt{3}$

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

One of the roots is $2-\sqrt{3}$ ,

Let the second root be $x$

Sum of roots $\frac{-b}{a}=-2(\sqrt{3}-1)=-2\sqrt{3}+2$

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

$x=-\sqrt{3}$