If root{3-2x} = root(2x) +1, then 4x^2 = : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 16 Jan 2017, 11:55

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92902 [0], given: 10528

If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

09 Jul 2012, 02:23
Expert's post
38
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

53% (03:03) correct 47% (02:03) wrong based on 1142 sessions

### HideShow timer Statistics

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

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

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92902 [4] , given: 10528

Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

09 Jul 2012, 02:23
4
KUDOS
Expert's post
11
This post was
BOOKMARKED
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: $$4x^2=6x-1$$.

_________________
Joined: 28 Mar 2012
Posts: 314
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 29

Kudos [?]: 398 [1] , given: 23

Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

09 Jul 2012, 05:16
1
KUDOS
1
This post was
BOOKMARKED
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

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

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.

Regards,
Manager
Affiliations: Project Management Professional (PMP)
Joined: 30 Jun 2011
Posts: 213
Location: New Delhi, India
Followers: 5

Kudos [?]: 67 [0], given: 12

Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

09 Jul 2012, 06:26
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

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

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.

Regards,

Hi

Can we solve equations and get the answer?
Like taking $$\sqrt{2x}$$ to right and squaring the both sides
_________________

Best
Vaibhav

If you found my contribution helpful, please click the +1 Kudos button on the left, Thanks

Joined: 28 Mar 2012
Posts: 314
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 29

Kudos [?]: 398 [0], given: 23

Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

09 Jul 2012, 20:48
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 $$4x^2=6x-1$$

Regards,

PS: But that isn't much fun.
Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92902 [1] , given: 10528

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

13 Jul 2012, 01:42
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
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: $$4x^2=6x-1$$.

_________________
Senior Manager
Joined: 23 Mar 2011
Posts: 473
Location: India
GPA: 2.5
WE: Operations (Hospitality and Tourism)
Followers: 19

Kudos [?]: 211 [0], given: 59

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

29 Apr 2013, 05:40
Bunuel wrote:
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

_________________

"When the going gets tough, the tough gets going!"

Bring ON SOME KUDOS MATES+++

-----------------------------

My GMAT journey begins: http://gmatclub.com/forum/my-gmat-journey-begins-122251.html

Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92902 [1] , given: 10528

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

29 Apr 2013, 05:56
1
KUDOS
Expert's post
sdas wrote:
Bunuel wrote:
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

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.
_________________
Senior Manager
Joined: 23 Mar 2011
Posts: 473
Location: India
GPA: 2.5
WE: Operations (Hospitality and Tourism)
Followers: 19

Kudos [?]: 211 [0], given: 59

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

29 Apr 2013, 07:15
Thanks Bunuel. I missed the last 2 steps....
_________________

"When the going gets tough, the tough gets going!"

Bring ON SOME KUDOS MATES+++

-----------------------------

My GMAT journey begins: http://gmatclub.com/forum/my-gmat-journey-begins-122251.html

Intern
Joined: 28 Jun 2011
Posts: 15
Followers: 0

Kudos [?]: 9 [0], given: 8

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

15 Dec 2013, 10:46
Hi guys,

can someone please explain why (root(2x) +1)^2 is not 2x +1 but 2x+1+2 root(2x). I thought I would just take every term to the secound power, such as root(2x)^2 + 1^2...

Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92902 [1] , given: 10528

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

16 Dec 2013, 00:18
1
KUDOS
Expert's post
steilbergauf wrote:
Hi guys,

can someone please explain why (root(2x) +1)^2 is not 2x +1 but 2x+1+2 root(2x). I thought I would just take every term to the secound power, such as root(2x)^2 + 1^2...

Must know property: $$(a+b)^2=a^2+2ab+b^2$$.

Check the following link for more: algebra-101576.html

Hope this helps.
_________________
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1858
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 47

Kudos [?]: 1926 [0], given: 193

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

17 Apr 2014, 02:20
Bunuel wrote:
If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

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

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600

$$\sqrt{3-2x} = \sqrt{2x} +1$$

$$\sqrt{3-2x} - \sqrt{2x} = 1$$

Squaring both sides

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

$$3 + \sqrt{24x - 16x^2} = 1$$

$$24x - 16x^2 = 4$$

$$16x^2 = 24x - 4$$

$$4x^2 = 6x - 1$$

_________________

Kindly press "+1 Kudos" to appreciate

Intern
Joined: 05 Feb 2014
Posts: 12
Followers: 0

Kudos [?]: 4 [0], given: 22

If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

31 Jul 2014, 03:31
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.

Thanks for any help!
Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92902 [0], given: 10528

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

31 Jul 2014, 06:23
pipe19 wrote:
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.

Thanks for any help!

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
_________________
Intern
Joined: 29 Oct 2014
Posts: 25
Followers: 0

Kudos [?]: 9 [0], given: 14

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

29 Oct 2014, 22:47
Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

Thanks a mil!
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1858
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 47

Kudos [?]: 1926 [0], given: 193

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

29 Oct 2014, 23:05
ColdSushi wrote:
Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

Thanks a mil!

$$(\sqrt{3-2x})^2 = 3-2x$$

$$(1-2x)^2 = (1-2x)(1-2x)$$

$$(x-y)^2 = x^2 - 2xy + y^2$$

$$x^2 - y^2 = (x+y)(x-y)$$

Hope this clarifies
_________________

Kindly press "+1 Kudos" to appreciate

Intern
Joined: 29 Oct 2014
Posts: 25
Followers: 0

Kudos [?]: 9 [0], given: 14

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

30 Oct 2014, 03:02
PareshGmat wrote:
ColdSushi wrote:
Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

Thanks a mil!

$$(\sqrt{3-2x})^2 = 3-2x$$

$$(1-2x)^2 = (1-2x)(1-2x)$$

$$(x-y)^2 = x^2 - 2xy + y^2$$

$$x^2 - y^2 = (x+y)(x-y)$$

Hope this clarifies

Wow - I totally mixed up the rules - thanks for clarifying.
Intern
Joined: 22 Oct 2014
Posts: 28
Followers: 0

Kudos [?]: 7 [0], given: 79

If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

15 Jan 2015, 07:16
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

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

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.

Regards,

Hi,
Could you please explain how you got into this 0<x<1 from the LHS and RHS?
Also, the calculations of highlighted portion in the answer choice E.
TIA
Intern
Joined: 09 Jan 2014
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 12

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

08 Feb 2015, 05:41
1
This post was
BOOKMARKED
Bunuel wrote:
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: $$4x^2=6x-1$$.

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.
Director
Joined: 10 Mar 2013
Posts: 608
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Followers: 15

Kudos [?]: 266 [0], given: 200

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

### Show Tags

04 May 2015, 11:51
see screenshot below
Attachments

Math.jpg [ 51.45 KiB | Viewed 8330 times ]

_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Re: If root{3-2x} = root(2x) +1, then 4x^2 =   [#permalink] 04 May 2015, 11:51

Go to page    1   2    Next  [ 26 posts ]

Similar topics Replies Last post
Similar
Topics:
4 If x+(1/x)=4, x2+(1/x2)=? 5 24 Sep 2016, 21:46
9 If m=9^(x−1), then in terms of m, 3^(4x−2) 4 03 Jun 2014, 13:20
2 If root{3-2x} = root(2x) +1, then 4x^2 = 6 05 Apr 2011, 19:34
15 If rot{3-2x} = root(2x) +1, then 4x^2 = 10 17 Jan 2011, 20:51
50 If x>0 then 1/[(root(2x)+root(x)]=? 24 14 Dec 2010, 18:08
Display posts from previous: Sort by