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06 Jan 2013, 04:42
Kudos
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Hey Guys,
as I have found this forum today I would like to share my thoughts on the GMAT with you. I´ve began preparing for the test about a week ago and right now I´m still fighting easy problems.
I got a qestion that needs two equations two be created. I got the first equation, which is:
\sqrt{3-2x} = \sqrt{2x}+1
When I square both sides, I do get
3-2x = ?
My problem is that I don´t know any rule that leads to the correct term on the right side, which Is (as indicated in the explanations)
3-2x = 2x+2*\sqrt{2x}+1
Anyone any idea how this works?
Maybe some of you can help me with this.
Thanks in advance
Max
as I have found this forum today I would like to share my thoughts on the GMAT with you. I´ve began preparing for the test about a week ago and right now I´m still fighting easy problems.
I got a qestion that needs two equations two be created. I got the first equation, which is:
\sqrt{3-2x} = \sqrt{2x}+1
When I square both sides, I do get
3-2x = ?
My problem is that I don´t know any rule that leads to the correct term on the right side, which Is (as indicated in the explanations)
3-2x = 2x+2*\sqrt{2x}+1
Anyone any idea how this works?
Maybe some of you can help me with this.
Thanks in advance
Max
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06 Jan 2013, 06:02
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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!
\(\sqrt{3-2x} = \sqrt{2x} -1\)
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!
06 Jan 2013, 06:11
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maxkay47
3 - 2x =( \sqrt{2x} + 1) * ( \sqrt{2x} + 1)
= 2x + \sqrt{2x} + \sqrt{2x} + 1
= 2x + 2 \sqrt{2x} + 1
