If $x^2 - x \sqrt{2} + 3x\sqrt{3}=\sqrt{54}$ then $x =$

The moment we see something monstrously complicated as

$x^2 - x \sqrt{2} + 3x\sqrt{3}=\sqrt{54}$

we can be sure that there is a simple solution.

The reason being, the GRE is not a test of our calculating prowess, but a test of our alertness in spotting opportunities that enable an easy solution for the problem. The GRE rewards clever test takers by providing them opportunities to save precious time.

Right off the bat, we see that it is a quadratic equation.

so, let us begin by gathering all the like terms and writing the equation in the standard form.

$x^2 +(3\sqrt{3} - \sqrt{2}) - \sqrt{54} = 0$

Now, we need two numbers which when added together give us the coefficient of $\text{x}$ and when multiplied together give us the constant term.

But we see that the coefficient of $\text{x}$ is already a sum, consisting of the following two numbers :

$3\sqrt{3}, -\sqrt{2}.$

Could it be that the product of these two numbers yields$-\sqrt{54}\text{ ? }$

Well, lets see. $3\sqrt{3} \times \sqrt{2} = -3\sqrt{6}$

And $-\sqrt{54} = -1 \times \sqrt{9 \times 6} = -3\sqrt{6 }$

We see that it is indeed the case. The two numbers are $3\sqrt{3} \text{ and } -\sqrt{2}$

Therefore,$x = -3\sqrt{3} \text{ and } \sqrt{2}$

The answers are Choices C and D