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M01-03

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M01-03 [#permalink]

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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. \(1\)
[Reveal] Spoiler: OA

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Re M01-03 [#permalink]

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New post 15 Sep 2014, 23:14
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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I understand why Choice B is a solution, but I don't understand why 0 - Choice A - is not..

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Re: M01-03 [#permalink]

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New post 16 Oct 2016, 21:40
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Re: M01-03 [#permalink]

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New post 17 Oct 2016, 06:47
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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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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Re: M01-03 [#permalink]

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New post 26 Jan 2017, 05:49
BoomHH wrote:
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 \(x\): \(\frac{1}{2}*\frac{1}{2} = 2*\frac{1}{4}\)."


Thank you. Edited.
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Collection of Questions:
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Re: M01-03 [#permalink]

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New post 26 Aug 2017, 16:07
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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Re: M01-03 [#permalink]

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New post 27 Aug 2017, 03:03
alainca wrote:
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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Collection of Questions:
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Re: M01-03 [#permalink]

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New post 21 Sep 2017, 11:31
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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Re: M01-03 [#permalink]

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New post 21 Sep 2017, 11:35
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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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M01-03 [#permalink]

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New post 21 Sep 2017, 12:04
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 :)
Re: M01-03   [#permalink] 21 Sep 2017, 12:04
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