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If rot{32x} = root(2x) +1, then 4x^2 =
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17 Jan 2011, 21:51
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If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) = (A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 OPEN DISCUSSION OF THIS QUESTION IS HERE: ifroot32xroot2x1then4x135539.html
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Re: OG Diag #16 PS
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18 Jan 2011, 02:04
tonebeeze wrote: This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types? If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) = (A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 Hey, below mentioned is my strategy to solve this problem, Sqrt both side you will get, 32x= 2x+1+2sqrt{2x} 32x2x1= 2sqrt{2x} 24x= 2sqrt{2x} devide by 2 both side 12x = sqrt{2x} Again sqrt of both side 1+4x^24x= 2x 1+4x^26x=0 4x^2= 6x1 answer is E.



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Re: OG Diag #16 PS
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18 Jan 2011, 15:32
SoniaSaini wrote: tonebeeze wrote: This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types? If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) = (A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 Hey, below mentioned is my strategy to solve this problem, Sqrt both side you will get, 32x= 2x+1+2sqrt{2x} 32x2x1= 2sqrt{2x} 24x= 2sqrt{2x} devide by 2 both side 12x = sqrt{2x} Again sqrt of both side 1+4x^24x= 2x 1+4x^26x=0 4x^2= 6x1 answer is E. Same way to solve it (about 2min).



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Re: OG Diag #16 PS
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18 Feb 2011, 15:14
can someone explain me this stage plz: Sqrt both side you will get, 32x= 2x+1+2sqrt{2x} thanks.
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Re: OG Diag #16 PS
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18 Feb 2011, 16:00



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Re: OG Diag #16 PS
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18 Feb 2011, 22:57
thanks. ur explanations is super clear. +1
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Re: If rot{32x} = root(2x) +1, then 4x^2 =
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25 Jun 2012, 13:05
Solving the equation, I arrive to this point: \(2  2\sqrt{2x}\)\(=\)\(4x\) However, there are more than one way to get \(4x^2\). For example: We multiply both sides by \(x\): \(2x  2x\sqrt{2x}\)\(=\)\(4x^2\) So, the answer not necessarily is \(6x 1\) How can we know how to manipulate the equation in order to arrive to one of the choices and not another valid solution?
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Re: If rot{32x} = root(2x) +1, then 4x^2 =
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26 Jun 2012, 03:37
metallicafan wrote: Solving the equation, I arrive to this point: \(2  2\sqrt{2x}\)\(=\)\(4x\)
How can we know how to manipulate the equation in order to arrive to one of the choices and not another valid solution? After this point, you see that none of the options has a square root sign. You need to get rid of it to get to the answer. Hence, it is apparent that you need to square it again. You then take the square root term to one side and the rest of the terms to the other side and square both sides of the equation. This helps you get rid of the square root sign.
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Re: If rot{32x} = root(2x) +1, then 4x^2 =
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10 May 2013, 10:16
wouldnt [sqr(32x)]^2 give us +/ (32x)????



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Re: If rot{32x} = root(2x) +1, then 4x^2 =
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11 May 2013, 03:14



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Re: If rot{32x} = root(2x) +1, then 4x^2 =
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11 May 2013, 04:43
mokura wrote: wouldnt [sqr(32x)]^2 give us +/ (32x)???? And one more thing, x^2 = 4 implies x = +2 or 2
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Re: If rot{32x} = root(2x) +1, then 4x^2 =
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