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M01-03

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M01-03  [#permalink]

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New post 16 Sep 2014, 00:14
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

83% (00:57) correct 17% (00:53) wrong based on 188 sessions

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Re M01-03  [#permalink]

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New post 16 Sep 2014, 00:14
Official Solution:

If \(x\) is a positive number and \(\frac{1}{2}*\sqrt{x} = 2x\), what is the value of \(x\)?

A. \(0\)
B. \(\frac{1}{16}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(1\)


Backsolve. Plug answer (C) into the equation for \(x\): \(\frac{1}{2}*\frac{1}{2} = 2*\frac{1}{4}\). Solve both sides of the equation: \(\frac{1}{4} \neq \frac{1}{2}\). Continue plugging in the answers. Plugging answer (B) into the equation for \(x\) gives: \(\frac{1}{2}*\frac{1}{4} = 2*\frac{1}{16}\)or \(\frac{1}{8} = \frac{1}{8}.\)

Another option is to solve the equation for \(x\). \(\frac{1}{2}*\sqrt{x} = 2x\). Since \(x\) is a positive number, i.e., \(x \gt 0\), square both sides of equation to get:
\(\frac{1}{4} * x = 4x^2\)

Multiply both sides by 4:
\(x = 16x^2\)
\(16x^2 - x = 0\)
\(x(16x-1) = 0\)

The roots of this equation are \(x = 0\) and \(x = \frac{1}{16}\). Since \(x \gt 0\), the correct answer is \(x = \frac{1}{16}\).


Answer: B
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Re: M01-03  [#permalink]

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New post 16 Oct 2016, 18:59
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I understand why Choice B is a solution, but I don't understand why 0 - Choice A - is not..

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New post 16 Oct 2016, 22:40
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Re: M01-03  [#permalink]

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New post 17 Oct 2016, 07:47
Bunuel wrote:
moulayabdesslam wrote:
I understand why Choice B is a solution, but I don't understand why 0 - Choice A - is not..

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The stem says that x is a positive number and 0 is neither positive nor negative.


Thanks! That's funny how one can miss such obvious information.

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Re: M01-03  [#permalink]

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New post 26 Jan 2017, 04:22
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Bunuel

I think there is a little Typo:

Backsolve. Plug answer (C) into the equation for \(x\): \(\frac{1}{2}*\frac{1}{2} = 2*\frac{1}{2}\).

If you plug-in answer (C) it should be "Backsolve. Plug answer (C) into the equation for \(x\): \(\frac{1}{2}*\frac{1}{2} = 2*\frac{1}{4}\)."
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New post 26 Jan 2017, 06:49
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Re: M01-03  [#permalink]

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New post 26 Aug 2017, 17:07
Instead of going from 1/4∗x=4x^2 and multiplying both sides by 4 and solving for roots, why not simply divide both sides by x, since you know x is positive?

You would then get 1/4 = 4x, then x = 1/16.
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New post 27 Aug 2017, 04:03
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New post 21 Sep 2017, 12:31
I am a little confused as to which one is the correct answer.

Is it now 1/4 or 1/16 ?

a) If you move 1/2 to RHS, square both sides and solve for x, then we would get 1/4.
b) Instead, if we square both LHS and RHS as they are, then the answer would be 1/16.

Can you please clarify? Thanks.
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New post 21 Sep 2017, 12:35
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theHermes5 wrote:
I am a little confused as to which one is the correct answer.

Is it now 1/4 or 1/16 ?

a) If you move 1/2 to RHS, square both sides and solve for x, then we would get 1/4.
b) Instead, if we square both LHS and RHS as they are, then the answer would be 1/16.

Can you please clarify? Thanks.


The correct answer is B: 1/16. You can see it under the spoiler in the first post as well as in the solution. How are you getting 1/4?
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Re: M01-03  [#permalink]

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New post 21 Sep 2017, 13:04
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Sorry, I made a rookie mistake. I missed squaring 4 when squaring both sides of the equation and ended up with 1/4.
1/16 is correct, like you mentioned :)
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New post 07 Jul 2018, 18:04
what about the negatives?
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New post 08 Jul 2018, 00:56
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New post 19 Feb 2019, 04:40
simplify the equation and then go by options
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Re: M01-03   [#permalink] 19 Feb 2019, 04:40
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