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If \(\sqrt{3 - x} = 1 -\sqrt{ 2x}, \) then \(9x^2 =\)
(A) 5
(B) 8
(C) 2 - 3x
(D) 12 - 4x
(E) 20x - 4

Source: Total GMAT Math - Jeff Sackmann


Can you please check whether you copied the question correctly? \(\sqrt{3 - x} = 1 -\sqrt{ 2x}\) has no solution.

By the way, this question is a copy of the following OG question: https://gmatclub.com/forum/if-3-2x-1-2- ... 35539.html

I believe the question is correctly copied from the source :angel: . Below is the explanation stated in the source:

EXPL: It’s far from clear at the outset how you’ll find the value of \(9x^2\), but it’s usually the case that you have to get rid of the radical signs to accomplish anything. You can start down that path by squaring both sides of the equation:

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

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

\(3x - 2 = 2\sqrt{2x} \)

There’s still a radical, so square both sides again:

\((2\sqrt{2x})^2 = (3x - 2)^2\)

\(4(2x) = 9x^2 - 12x + 4\)

\(9x^2 = 20x - 4\), choice (E).

[end of explaination]

Thanks for providing the OG reference :please:. I am starting to gather that a lot of the Total GMAT Challenge questions are some variation of the harder OG questions

The problem there is that the equation in the original question, \(\sqrt{3-2x} = \sqrt{2x} +1\), has a solution: \(x = \frac{3}{4} - \frac{\sqrt{5}}{4}\), while the equation in this question, \(\sqrt{3 - x} = 1 -\sqrt{ 2x}\), has no solution for x. Proper GMAT question won't give you an equation that has no real solujtion..
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3-x=1+2x-2rt(2x)
3x-2=2rt(2x)
9x^2+4-12x=8x
=>9x^2=20x-4

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