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If x > 0 and 2x - 1 = 1/(2x + 1), then x =

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Bunuel
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If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink] New post  30 Jun 2019, 23:41
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If \(x > 0\) and \(2x-1=\frac{1}{2x+1}\), then \(x=\)


A. \(\frac{1}{2}\)

B. \(\frac{\sqrt{2}}{2}\)

C. \(1\)

D. \(\sqrt{2}\)

E. \(\sqrt{2} + 1\)
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If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink] New post  30 Jun 2019, 23:59
Bunuel wrote:
If \(x > 0\) and \(2x-1=\frac{1}{2x+1}\), then \(x=\)


A. \(\frac{1}{2}\)

B. \(\frac{\sqrt{2}}{2}\)

C. \(1\)

D. \(\sqrt{2}\)

E. \(\sqrt{2} + 1\)


The expression when cross multiplied is in the form a^2-b^2=(a-b)(a+b); So 4x^2-1=1
x^2=1/2; x=1/root2 ; multiply the numerator and denominator by root2 to get Option B
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Re: If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink] New post  01 Jul 2019, 00:11
Constraint x>0
2x-1 = 1/(2x+1)

(2x-1)(2x+1) = 1
((2x)^2-1) =1
(2x)^2 = 2 —> x= rt2/2

Answer B

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If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink] New post  01 Jul 2019, 00:27
Bunuel wrote:
If \(x > 0\) and \(2x-1=\frac{1}{2x+1}\), then \(x=\)


A. \(\frac{1}{2}\)

B. \(\frac{\sqrt{2}}{2}\)

C. \(1\)

D. \(\sqrt{2}\)

E. \(\sqrt{2} + 1\)


(2x-1)(2x+1) = 1
\(4x^2 -1 = 1\)
\(4x^2 = 2\)
\(x^2 = 2 / 4\)
\(|x| = \sqrt{2}/2\)
\(x = \sqrt{2}/2\) since x>0


IMO B
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Re: If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink] New post  01 Jul 2019, 00:37
Bunuel wrote:
If \(x > 0\) and \(2x-1=\frac{1}{2x+1}\), then \(x=\)


A. \(\frac{1}{2}\)

B. \(\frac{\sqrt{2}}{2}\)

C. \(1\)

D. \(\sqrt{2}\)

E. \(\sqrt{2} + 1\)



rearranging the second equation

\((2x)^2-1^2=1\)

\(4x^2=2\)

\(x^2=1/2\)

\(x=1/\sqrt{2}\)

\(x=\sqrt{2}/2\)
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Re: If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink] New post  01 Jul 2019, 01:38
Bunuel wrote:
If \(x > 0\) and \(2x-1=\frac{1}{2x+1}\), then \(x=\)


A. \(\frac{1}{2}\)

B. \(\frac{\sqrt{2}}{2}\)

C. \(1\)

D. \(\sqrt{2}\)

E. \(\sqrt{2} + 1\)


\(2x-1=\frac{1}{2x+1}\)
--> \((2x-1)(2x+1) = 1\)
--> \(4x^2 - 1 = 1\)
--> \(4x^2 = 2\)
--> \(2x = ±\sqrt{2}\) = \(\sqrt{2}\) {Since, x > 0}
--> x = \(\sqrt{2}/2\)

IMO Option B

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Re: If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink] New post  01 Jul 2019, 07:41
simplify
\(2x-1=\frac{1}{2x+1}\), then \(x=\)
we get
4x^2-1=1
solve for x= \(\frac{\sqrt{2}}{2}\)
IMO B

Bunuel wrote:
If \(x > 0\) and \(2x-1=\frac{1}{2x+1}\), then \(x=\)


A. \(\frac{1}{2}\)

B. \(\frac{\sqrt{2}}{2}\)

C. \(1\)

D. \(\sqrt{2}\)

E. \(\sqrt{2} + 1\)
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Re: If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink] New post  10 Sep 2021, 04:53
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x > 0 and 2x - 1 = 1/(2x + 1), then x = [#permalink]
10 Sep 2021, 04:53
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