May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55272

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
30 May 2015, 04:43
reto wrote: Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) = (A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 Diagnostic Test Question: 16 Page: 22 Difficulty: 600 Why is this question categorized as difficulty 700 above and 600 in the question steam? What's the actual difficulty? If you square the expression \(\sqrt{2x} +1\) and there are NO paranthesis, why is it not correct to square each term seperately to get 2x + 1?` Thank you so much 1. 600 difficulty was given when posted. 700 difficulty is based on users' answers on the question. 2. (a + b)^2 generally does not equal to a^2 + b^2. I'ts pretty basic actually.
_________________



Retired Moderator
Joined: 29 Apr 2015
Posts: 837
Location: Switzerland
Concentration: Economics, Finance
WE: Asset Management (Investment Banking)

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
30 May 2015, 05:04
Bunuel wrote: reto wrote: Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) = (A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 Diagnostic Test Question: 16 Page: 22 Difficulty: 600 Why is this question categorized as difficulty 700 above and 600 in the question steam? What's the actual difficulty? If you square the expression \(\sqrt{2x} +1\) and there are NO paranthesis, why is it not correct to square each term seperately to get 2x + 1?` Thank you so much 1. 600 difficulty was given when posted. 700 difficulty is based on users' answers on the question. 2. (a + b)^2 generally does not equal to a^2 + b^2. I'ts pretty basic actually. Thanks. Thats what I meant in the question there are no paranthesis, but when you squared it in your solution, you gave the expression extra paranthesis  why? I know it's not the same. But I don't understand why you completed it when squaring...
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!
PS Please send me PM if I do not respond to your question within 24 hours.



Math Expert
Joined: 02 Sep 2009
Posts: 55272

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
30 May 2015, 05:21
reto wrote: Thanks. Thats what I meant in the question there are no paranthesis, but when you squared it in your solution, you gave the expression extra paranthesis  why? I know it's not the same. But I don't understand why you completed it when squaring... Squaring means multiplying by the same number, so when you square a + b, you get (a + b)(a + b) = (a + b)^2. I'd advice to brush up fundamentals before practicing questions, especially, hard ones.
_________________



Board of Directors
Joined: 17 Jul 2014
Posts: 2551
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
20 Mar 2016, 18:25
Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) = (A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 Diagnostic Test Question: 16 Page: 22 Difficulty: 600 sqrt(32x)1 = sqrt(2x)  square everything 2x = 32x  2*sqrt(32x)+1 2x=4  2x  2*sqrt(32x) 44x=2*sqrt(32x)  square everything again 16+16x^2 32x = 12  8x  add 8x to both sides, subtract 12 from both sides 16x^2  24x +4 = 0  divide by 4 4x^2 6x +1 = 0  add 6x to both sides and subtract 1: 4x^2 = 6x1 E



Director
Joined: 17 Dec 2012
Posts: 630
Location: India

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
08 Mar 2018, 18:48
Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) = (A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 Diagnostic Test Question: 16 Page: 22 Difficulty: 600 Main Idea:While simplifying keep the more complex root expression alone on one side and simplify. Details: sqrt(32x)=sqrt(2) +1. Squaring on both sides: 32x=2x+1+2*sqrt(2x) Simplifying we get (4x)^2 = 6x1 Hence E.
_________________
Srinivasan Vaidyaraman Sravna Test Prep http://www.sravnatestprep.comHolistic and Systematic Approach



VP
Joined: 09 Mar 2016
Posts: 1283

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
02 May 2018, 08:51
Bunuel wrote: SOLUTION
If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
\(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\).
Answer: E. Bunuel can you please share your expertise knowledge on the following how from this \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) we get this \(32x=2x+2*\sqrt{2x}+1\) ? thank you for your kind attention to my question



Math Expert
Joined: 02 Sep 2009
Posts: 55272

If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
02 May 2018, 09:17
dave13 wrote: Bunuel wrote: SOLUTION
If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
\(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\).
Answer: E. Bunuel can you please share your expertise knowledge on the following how from this \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) we get this \(32x=2x+2*\sqrt{2x}+1\) ? thank you for your kind attention to my question Let me asks you: 1. What does \((\sqrt{a})^2\) equal to? 2. What does \((a + b)^2\) equal to? If you can answer these questions, you should be able to figure out your doubt.
_________________



VP
Joined: 09 Mar 2016
Posts: 1283

If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
02 May 2018, 10:36
Bunuel wrote: dave13 wrote: Bunuel wrote: SOLUTION
If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
\(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\).
Answer: E. Bunuel can you please share your expertise knowledge on the following how from this \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) we get this \(32x=2x+2*\sqrt{2x}+1\) ? thank you for your kind attention to my question Let me asks you: 1. What does \((\sqrt{a})^2\) equal to? 2. What does \((a + b)^2\) equal to? If you can answer these questions, you should be able to figure out your doubt. Hi Bunuel thanks for your questions, i will gladly answer your questions:) 1. What does \((\sqrt{a})^2\) equal to? it equals \(a\) 2. What does \((a + b)^2\) equal to? it equals this \((a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2\) now let me get back to the question \((\sqrt{32x})^2 = (\sqrt{2x} +1)^2\) square both sides (using this formula \((\sqrt{a})^2\) = \(a\) so we get \((32x)^2 = (2x1)^2\) now using this formula \((a + b)^2\) = \(a^2+2ab+b^2\) we get \((32x) ( 32x) = (2x+1) (2x+1)\) so from here we get \(10=16x\):? can you please explain what am i doing wrong knowing formula is one thing, but applying it correctly is quite challenging for me ......sometimes



Math Expert
Joined: 02 Sep 2009
Posts: 55272

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
02 May 2018, 12:28
dave13 wrote: Bunuel wrote: dave13 wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 \(\sqrt{32x} = \sqrt{2x} +1\) > square both sides: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) > \(32x=2x+2*\sqrt{2x}+1\) > rearrange so that to have root at one side: \(24x=2*\sqrt{2x}\) > reduce by 2: \(12x=\sqrt{2x}\) > square again: \((12x)^2=(\sqrt{2x})^2\) > \(14x+4x^2=2x\) > rearrange again: \(4x^2=6x1\). Answer: E. Bunuel can you please share your expertise knowledge on the following how from this \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) we get this \(32x=2x+2*\sqrt{2x}+1\) ? thank you for your kind attention to my question Let me asks you: 1. What does \((\sqrt{a})^2\) equal to? 2. What does \((a + b)^2\) equal to? If you can answer these questions, you should be able to figure out your doubt. Hi Bunuel thanks for your questions, i will gladly answer your questions:) 1. What does \((\sqrt{a})^2\) equal to? it equals \(a\) 2. What does \((a + b)^2\) equal to? it equals this \((a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2\) now let me get back to the question \((\sqrt{32x})^2 = (\sqrt{2x} +1)^2\) square both sides (using this formula \((\sqrt{a})^2\) = \(a\) so we get \((32x)^2 = (2x1)^2\) now using this formula \((a + b)^2\) = \(a^2+2ab+b^2\) we get \((32x) ( 32x) = (2x+1) (2x+1)\) so from here we get \(10=16x\):? can you please explain what am i doing wrong knowing formula is one thing, but applying it correctly is quite challenging for me ......sometimes 1. Yes, \((\sqrt{a})^2=a\), hence \((\sqrt{32x})^2 =32x\). 2. Here you are also correct: \((a + b)^2=a^2+2ab+b^2\), hence \((\sqrt{2x} +1)^2=(\sqrt{2x})^2 + 2*\sqrt{2x}*1+1^2=2x+2*\sqrt{2x}+1\). Does this make sense?
_________________



VP
Joined: 09 Mar 2016
Posts: 1283

If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
03 May 2018, 10:31
Bunuel, it makes perfect sense ! i applied this formula \((a  b)^2\) = \(a^22ab+b^2\) to \((12x)^2=(\sqrt{2x})^2\) so i get this \(1^2  2 * 1 *(2x) +2x^2 = 2x\) > \(1 22x+4x=2x\) > \(1 2+2x= 2x\) > \(1 =0\):? what am i doing wrong , can you please explain in two or three words many thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 55272

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
03 May 2018, 11:16
dave13 wrote: Bunuel, it makes perfect sense ! i applied this formula \((a  b)^2\) = \(a^22ab+b^2\) to \((12x)^2=(\sqrt{2x})^2\) so i get this 1^2  2 * 1 * (2x) + 2x^2 = 2x > \(1 22x+4x=2x\) > \(1 2+2x= 2x\) > \(1 =0\):? what am i doing wrong , can you please explain in two or three words many thanks! I highlighted the mistakes. \((1  2x)^2 = 1  2*2x + (2x)^2 = 1  4x + 4x^2\)
_________________



VP
Joined: 09 Mar 2016
Posts: 1283

If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
03 May 2018, 12:19
Bunuel wrote: dave13 wrote: Bunuel, it makes perfect sense ! i applied this formula \((a  b)^2\) = \(a^22ab+b^2\) to \((12x)^2=(\sqrt{2x})^2\) so i get this 1^2  2 * 1 * (2x) + 2x^2 = 2x > \(1 22x+4x=2x\) > \(1 2+2x= 2x\) > \(1 =0\):? what am i doing wrong , can you please explain in two or three words many thanks! I highlighted the mistakes. \((1  2x)^2 = 1  2*2x + (2x)^2 = 1  4x + 4x^2\) Bunuel thank you ! just last question how did you convert this \((2x)^2\) and got \(4x^2\) \((2x)^2\) equals \(4x\) and not \(4x^2\) please explain this part and I will be more than happy



Math Expert
Joined: 02 Sep 2009
Posts: 55272

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
03 May 2018, 12:21
dave13 wrote: Bunuel wrote: dave13 wrote: Bunuel, it makes perfect sense ! i applied this formula \((a  b)^2\) = \(a^22ab+b^2\) to \((12x)^2=(\sqrt{2x})^2\) so i get this 1^2  2 * 1 * (2x) + 2x^2 = 2x > \(1 22x+4x=2x\) > \(1 2+2x= 2x\) > \(1 =0\):? what am i doing wrong , can you please explain in two or three words many thanks! I highlighted the mistakes. \((1  2x)^2 = 1  2*2x + (2x)^2 = 1  4x + 4x^2\) Bunuel thank you ! just last question how did you convert this \((2x)^2\) and got \(4x^2\) \((2x)^2\) equals \(4x\)and not \(4x^2\) please explain this part and I will be more than happy We have \((2x)^2\). Notice that we are squaring 2x, so you should square both 2 and x: \((2x)^2=2x*2x=4x^2\)
_________________



Director
Joined: 29 Jun 2017
Posts: 707

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
14 Feb 2019, 09:58
this is tricky the key here is to arrange the equation so that we have root at one side. when this happen, we can square and after squaring, we can have simple result. we square the first time immediately because the left side is a root already after getting the result fo the first squaring, we arrange so that we have root at one side.
look at the way Bunnuel do. I just explain more.



VP
Joined: 09 Mar 2018
Posts: 1004
Location: India

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
14 Feb 2019, 10:16
Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) = (A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 Diagnostic Test Question: 16 Page: 22 Difficulty: 600 \(\sqrt{32x} = \sqrt{2x} +1\) Square once to get 32x = 2x + 1 + 2\(\sqrt{2x}\) 12x = \(\sqrt{2x}\) Square again to get \((12x)^2\) = 2x 4x^2 = 6x  1 E
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.
Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up.



CEO
Joined: 12 Sep 2015
Posts: 3726
Location: Canada

Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
Show Tags
18 Mar 2019, 10:07
Bunuel wrote: If \(\sqrt{32x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
GIVEN: \(\sqrt{32x} = \sqrt{2x} +1\) Square both sides to get: \((\sqrt{32x})^2 =(\sqrt{2x} +1)^2\) Expand and simplify to get: \(32x=2x+2\sqrt{2x}+1\) Subtract 1 from both sides to get: \(12x=2x+2\sqrt{2x}\) Subtract 2x from both sides to get: \(24x=2\sqrt{2x}\) Divide both sides by 2 to get: \(12x=\sqrt{2x}\) Square both sides to get: \((12x)^2=(\sqrt{2x})^2\) Expand and simplify to get: \(14x+4x^2=2x\) Add 4x to both sides: \(1 + 4x^2=6x\) Subtract 1 from both sides to get: \(4x^2=6x1\) Answer: E Cheers, Brent
_________________
Test confidently with gmatprepnow.com




Re: If root{32x} = root(2x) +1, then 4x^2 =
[#permalink]
18 Mar 2019, 10:07



Go to page
Previous
1 2
[ 36 posts ]



