Last visit was: 19 Nov 2025, 14:44 It is currently 19 Nov 2025, 14:44
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
705-805 Level|   Algebra|   Roots|                        
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
 [535]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,361
 [535]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 09 Jul 2012, 03:23
29
Kudos
Add Kudos
503
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

59% (02:27) correct 41% (02:36) wrong based on 7688 sessions

History

Date
Time
Result
Not Attempted Yet
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
ShowHide Answer
Official Answer
E
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
 [122]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,361
 [122]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 09 Jul 2012, 03:23
47
Kudos
Add Kudos
75
Bookmarks
Bookmark this Post
SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

\(\sqrt{3-2x} = \sqrt{2x} +1\).

Square both sides: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\);

\(3-2x=2x+2*\sqrt{2x}+1\)

Rearrange so that to have root at one side: \(2-4x=2*\sqrt{2x}\);

Reduce by 2: \(1-2x=\sqrt{2x}\);

Square again: \((1-2x)^2=(\sqrt{2x})^2\);

\(1-4x+4x^2=2x\);

Rearrange again: \(4x^2=6x-1\).

Answer: E.
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,739
Own Kudos:
35,353
 [18]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,739
Kudos: 35,353
 [18]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = Updated on: 20 Jul 2020, 11:01
Originally posted by BrentGMATPrepNow on 18 Mar 2019, 10:07.
Last edited by BrentGMATPrepNow on 20 Jul 2020, 11:01, edited 2 times in total.
11
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
Show more

GIVEN: \(\sqrt{3-2x} = \sqrt{2x} +1\)

Square both sides to get: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\)

In other words: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)(\sqrt{2x} +1)\)

Use FOIL to expand right side: \((\sqrt{3-2x})^2 =(\sqrt{2x})^2 + 1\sqrt{2x} + 1\sqrt{2x} + 1^2\)

Simplify right side to get: \(3-2x=2x+2\sqrt{2x}+1\)

Subtract 1 from both sides to get: \(2-2x=2x+2\sqrt{2x}\)

Subtract 2x from both sides to get: \(2-4x=2\sqrt{2x}\)

Divide both sides by 2 to get: \(1-2x=\sqrt{2x}\)

Square both sides to get: \((1-2x)^2=(\sqrt{2x})^2\)

Expand and simplify to get: \(1-4x+4x^2=2x\)

Add 4x to both sides: \(1 + 4x^2=6x\)

Subtract 1 from both sides to get: \(4x^2=6x-1\)

Answer: E

Cheers,
Brent
General Discussion
User avatar
cyberjadugar
User avatar
Joined: 29 Mar 2012
Last visit: 28 May 2024
Posts: 265
Own Kudos:
1,755
 [12]
Given Kudos: 23
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
GMAT 3: 730 Q50 V38
Posts: 265
Kudos: 1,755
 [12]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 09 Jul 2012, 06:16
8
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
Show more
Hi,

Put x=1,
LHS= 1,
RHS=2.414
or LHS < RHS

Put x=0,
LHS= 1.732
RHS=1
LHS>RHS,

Thus, 0< x < 1, for equality to hold true.
or 0 < 2x < 2, squaring,
\(0 < 4x^2 < 4\)

Lets check the options now,
(A) 1 (\(0 < 4x^2 < 4\))
(B) 4 (\(0 < 4x^2 < 4\))
(C)2 − 2x (0< x < 1, thus, 0< 2 − 2x < 2, but \(0 < 4x^2 < 4\))
(D) 4x − 2 (0< x < 1, thus, -2< 4x-2 < 2, but \(0 < 4x^2 < 4\))
(E) 6x − 1 (0< x < 1, thus, -1< 6x-1 < 5, and \(0 < 4x^2 < 4\))

Thus, only possible option which lies in the range is E.

Answer (E),

Regards,
User avatar
cyberjadugar
User avatar
Joined: 29 Mar 2012
Last visit: 28 May 2024
Posts: 265
Own Kudos:
1,755
 [10]
Given Kudos: 23
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
GMAT 3: 730 Q50 V38
Posts: 265
Kudos: 1,755
 [10]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 09 Jul 2012, 21:48
5
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
Hi narangvaibhav,

Yes definitely, we can use the conventional method too,
\(\sqrt{3-2x} = \sqrt{2x} +1\)
Squaring both the sides,
\(3-2x=2x+1+2 \sqrt{2x}\)
or \(1-2x=\sqrt{2x}\)
Again, squaring both the sides,
\(1+4x^2-4x=2x\)
or \(4x^2=6x-1\)

Answer (E).

Regards,

PS: But that isn't much fun. :twisted:
User avatar
sdas
User avatar
Joined: 23 Mar 2011
Last visit: 06 May 2013
Posts: 365
Own Kudos:
661
 [1]
Given Kudos: 59
Location: India
Schools: Ivey '15 Queens'16 INSEAD Jan '16 Rotman '16
GPA: 2.5
WE:Operations (Hospitality and Tourism)
Schools: Ivey '15 Queens'16 INSEAD Jan '16 Rotman '16
Posts: 365
Kudos: 661
 [1]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 29 Apr 2013, 06:40
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Show more

Can someone help me solve this?

\(\sqrt{(3-2x)}\) = \(\sqrt{2x}\) + 1
\(\sqrt{(3-2x)}\) + 1 = \(\sqrt{2x}\)
If I square both sides,
2 - \(\sqrt{(3-2x)}\) = 2x
So, \(4x^2\) = 4 - (3-2x) - 2\(\sqrt{(3-2x)}\)

I am unable to reach to E

However, if I squared the main problem without shifting 1, then I am reaching to E

Please help
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
 [3]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,361
 [3]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 29 Apr 2013, 06:56
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sdas
Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1


Can someone help me solve this?

\(\sqrt{(3-2x)}\) = \(\sqrt{2x}\) + 1
\(\sqrt{(3-2x)}\) + 1 = \(\sqrt{2x}\)
If I square both sides,
2 - \(\sqrt{(3-2x)}\) = 2x
So, \(4x^2\) = 4 - (3-2x) - 2\(\sqrt{(3-2x)}\)

I am unable to reach to E

However, if I squared the main problem without shifting 1, then I am reaching to E

Please help
Show more

You could solve this way too:

\(\sqrt{3-2x} = \sqrt{2x} +1\)

Re-arrange: \(\sqrt{3-2x}-1 = \sqrt{2x}\);

Square: \((3-2x)-2\sqrt{3-2x}+1=2x\);

Re-arrange and reduce by 2: \(\sqrt{3-2x}=2-2x\);

Square: \(3-2x=4-8x+4x^2\)

\(4x^2=6x-1\).

Hope it's clear.
avatar
pipe19
avatar
Joined: 05 Feb 2014
Last visit: 19 Jun 2015
Posts: 11
Own Kudos:
11
 [2]
Given Kudos: 22
Posts: 11
Kudos: 11
 [2]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 31 Jul 2014, 04:31
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,

I have one question. I got to the equation 1 = 2x +\sqrt{2x} but then I squared 1- \sqrt{2x} = 2x to get to 4x^2 easily but I got stuck with the left term. So my question is how can I get the solution with this approach and additionally why did you guys square the term the other way round? It makes much more sense to me to square 2x in order to arrive at 4x^2 instead of squaring 1-2x.

I hope you understand what I'm asking. :wink:

Thanks for any help!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
 [3]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,361
 [3]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 31 Jul 2014, 07:23
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
pipe19
Hi,

I have one question. I got to the equation 1 = 2x +\sqrt{2x} but then I squared 1- \sqrt{2x} = 2x to get to 4x^2 easily but I got stuck with the left term. So my question is how can I get the solution with this approach and additionally why did you guys square the term the other way round? It makes much more sense to me to square 2x in order to arrive at 4x^2 instead of squaring 1-2x.

I hope you understand what I'm asking. :wink:

Thanks for any help!
Show more

First of all please read this:

rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum)

Next, if you square \(1-\sqrt{2x}=2x\) you'll get \(1 - 2\sqrt{2x}+2x=4x^2\). As you can see we don't have this answer among the options, so we should have done the other way around.

There are two ways of solving this given above:
if-root-3-2x-root-2x-1-then-4x-135539.html#p1104029
if-root-3-2x-root-2x-1-then-4x-135539.html#p1218265
avatar
melaos
avatar
Joined: 09 Jan 2014
Last visit: 19 Apr 2024
Posts: 5
Own Kudos:
1
Given Kudos: 12
Posts: 5
Kudos: 1
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 08 Feb 2015, 06:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) --> \(3-2x=2x+2*\sqrt{2x}+1\) --> rearrange so that to have root at one side: \(2-4x=2*\sqrt{2x}\) --> reduce by 2: \(1-2x=\sqrt{2x}\) --> square again: \((1-2x)^2=(\sqrt{2x})^2\) --> \(1-4x+4x^2=2x\) --> rearrange again: \(4x^2=6x-1\).

Answer: E.
Show more

sorry but a noob question here, i thought we aren't allowed to square both sides like this?

but instead all we can do is just to multiply both sides or divide both sides by the same number/objects etc.

cause when you square both sides, both sides is multiplying with different things right???

this part i don't get.
User avatar
EgmatQuantExpert
User avatar
e-GMAT Representative
Joined: 04 Jan 2015
Last visit: 02 Apr 2024
Posts: 3,663
Own Kudos:
20,167
 [2]
Given Kudos: 165
Expert
Expert reply
Posts: 3,663
Kudos: 20,167
 [2]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 05 May 2015, 07:19
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
melaos
Bunuel
SOLUTION

If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) --> \(3-2x=2x+2*\sqrt{2x}+1\) --> rearrange so that to have root at one side: \(2-4x=2*\sqrt{2x}\) --> reduce by 2: \(1-2x=\sqrt{2x}\) --> square again: \((1-2x)^2=(\sqrt{2x})^2\) --> \(1-4x+4x^2=2x\) --> rearrange again: \(4x^2=6x-1\).

Answer: E.

sorry but a noob question here, i thought we aren't allowed to square both sides like this?

but instead all we can do is just to multiply both sides or divide both sides by the same number/objects etc.

cause when you square both sides, both sides is multiplying with different things right???

this part i don't get.
Show more

Hi melaos,

You are right when you say that we need to multiply or divide by the same number/object on both sides of an equation. Let me tell you why squaring both sides of an equation is a perfectly acceptable way.

Consider an equation a = b. This equation tells us that the value of a is the same as value of b. When we square a number, we multiply the number by itself. So, squaring this equation would look like a*a = b* b. Now, you would be tempted to say that we are multiplying different numbers on both sides of the equation.

Before you say so, look back at our original equation which says a = b. Hence, when we multiply a on LHS and b on RHS we are multiplying both sides by variables which have the same values.
Hence, squaring both sides of an equation is allowed.

Hope its clear!

Regards
Harsh
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 19 Nov 2025
Posts: 5,794
Own Kudos:
5,511
Given Kudos: 161
Location: India
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
My Reviews
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
Posts: 5,794
Kudos: 5,511
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 14 Jan 2023, 02:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Asked: If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

\(3 - 2x = 2x + 1 + 2\sqrt{2x}\)
\(2 - 4x = 2\sqrt{2x}\)
\((1-2x)^2 = 4x^2 + 1 - 4x = 2x\)
\(4x^2 = 6x - 1\)

IMO E
User avatar
amitarya
User avatar
Joined: 25 Jul 2023
Last visit: 19 Nov 2025
Posts: 20
Own Kudos:
114
Given Kudos: 97
Products:
GMAT Club Tests
Posts: 20
Kudos: 114
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 02 Dec 2023, 01:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel.. Why u r not proceeding here with modulus while solving square root.
As square of (3-2x)^(1/2) should be written as mod(3-2x)..??

Please heldp
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
 [3]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,361
 [3]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 02 Dec 2023, 02:04
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
amitarya
Bunuel.. Why u r not proceeding here with modulus while solving square root.
As square of (3-2x)^(1/2) should be written as mod(3-2x)..??

Please heldp
Show more

\(\sqrt{a^2} = |a|\). Do we have the square of an expression in \(\sqrt{3-2x}\)? No. Hence, \(\sqrt{3-2x}\) does not equal to \(|3-2x|\). If it were \(\sqrt{(3-2x)^2}\), then it would be equal to \(|3-2x|\).
User avatar
amitarya
User avatar
Joined: 25 Jul 2023
Last visit: 19 Nov 2025
Posts: 20
Own Kudos:
114
Given Kudos: 97
Products:
GMAT Club Tests
Posts: 20
Kudos: 114
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 02 Dec 2023, 02:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
amitarya
Bunuel.. Why u r not proceeding here with modulus while solving square root.
As square of (3-2x)^(1/2) should be written as mod(3-2x)..??

Please heldp

\(\sqrt{a^2} = |a|\). Do we have the square of an expression in \(\sqrt{3-2x}\)? No. Hence, \(\sqrt{3-2x}\) does not equal to \(|3-2x|\). If it were \(\sqrt{(3-2x)^2}\), then it would be equal to \(|3-2x|\).
Show more


Bunuel.. Got it.. Thankyou so much for clarification..
You are always a saviour.. :)
:)
User avatar
Oppenheimer1945
User avatar
Joined: 16 Jul 2019
Last visit: 19 Nov 2025
Posts: 784
Own Kudos:
639
Given Kudos: 223
Location: India
Schools: INSEAD '26 (D) ISB '27 (D) IIMB '26 (A) IIMA '26 (A) HEC '26 (D) LBS '26 IESE '27 (D) Fuqua '27 (WL) Tepper '27 (WL)
GMAT Focus 1: 645 Q90 V76 DI80
GPA: 7.81
Schools: INSEAD '26 (D) ISB '27 (D) IIMB '26 (A) IIMA '26 (A) HEC '26 (D) LBS '26 IESE '27 (D) Fuqua '27 (WL) Tepper '27 (WL)
GMAT Focus 1: 645 Q90 V76 DI80
Posts: 784
Kudos: 639
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 21 Mar 2024, 23:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
3-2x=2x+1+2rt(2x)
2-4x=2rt(2x)
4+16x^2-16x=8x
16x^2=24x-4
4x^2=6x-1
User avatar
DanTheGMATMan
User avatar
Joined: 02 Oct 2015
Last visit: 18 Nov 2025
Posts: 378
Own Kudos:
227
 [1]
Given Kudos: 9
Expert
Expert reply
Posts: 378
Kudos: 227
 [1]
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 29 May 2024, 20:36
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
­Straighforward algebra stuff we love to do:

User avatar
ablatt4
User avatar
Joined: 18 Dec 2024
Last visit: 24 Sep 2025
Posts: 89
Own Kudos:
2
Given Kudos: 89
Location: United States (FL)
Concentration: Finance
Schools: Duke '26 Darden '26 Johnson '26 UNC '26
Schools: Duke '26 Darden '26 Johnson '26 UNC '26
Posts: 89
Kudos: 2
If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 30 Aug 2025, 17:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sqrt 3-2x = sqrt 2x +1
3-2x=2x+2sqrt 2x +1
0=4x+2sqrt 2x -2
0=2x + sqrt 2x -1
-2x+1 = sqrt 2x
4x^2-4x+1=2x
4x^2=6x-1

E

Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
Show more
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts