Question

In each of the following, determine whether the given values are solutions of the given equation or not :
$x^2-3\sqrt{3x}+6=0,x=\sqrt{3},x=-2\sqrt{3}$

• A. x = $\sqrt{3}$ is not a solution but x = -2$\sqrt{3}$ is a solution.
• B. x = $\sqrt{3}$ is a solution but x = -2$\sqrt{3}$ is not a solution.
  
• C. x = $\sqrt{3}$ & x = -2$\sqrt{3}$ are not solutions.
• D. x = $\sqrt{3}$ & x = -2$\sqrt{3}$ aresolutions.

Answer 1

Answer: (B)

x = $\sqrt{3}$ is a solution but x = -2$\sqrt{3}$ is not a solution.

$\Rightarrow$ $x^2-3\sqrt{3}x+6=0$

$\Rightarrow$ $x^2-2\sqrt{3}x-\sqrt{3}x+6=0$

$\Rightarrow$ $x(x-2\sqrt{3})-\sqrt{3}(x-2\sqrt{3})=0$

$\Rightarrow$ $(x-2\sqrt{3})(x-\sqrt{3})=0$

$\Rightarrow$ $x-2\sqrt{3}=0$ and $x-\sqrt{3}=0$

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

$\therefore$ Here, we can see correct answer is option B