Math question: Quite easy (maybe at least for you ;))

Hi there, Try squaring the equation one more time:

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

Squaring both sides:
\(3-2x = 2x - 2.\sqrt{2x} +1\)
=> \(\sqrt{2x} =2x-1\)
Squaring both sides once again:

\(2x = 4 x^{2} -4x + 1\)
=> \(4 x^{2} -6x +1 = 0\)
=> \(x =\frac{3+- \sqrt{5}}{4}\)

Now for the tricky part. Since you squared a couple of times the original equation you certainly changed the nature of the equation (meaning, you may have inserted extra solutions that aren't in the original equation). So, you have to test your solutions in the original equation to see which ones still aply.

For the solution \(k = \frac{3- \sqrt{5}}{4}\) you get that \(\sqrt{2k}-1\)is negative. Which is no good, since the left side of the original equation has to be always positive. The only solution that fits is


\(x = \frac{3+\sqrt{5}}{4}\\
\)
Cheers!