GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 25 Mar 2019, 21:16 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # If root{3-2x} = root(2x) +1, then 4x^2 =

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 53831
If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

14
77 00:00

Difficulty:   95% (hard)

Question Stats: 52% (02:25) correct 48% (02:25) wrong based on 1620 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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 53831
Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

18
22
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: $$4x^2=6x-1$$.

_________________
##### General Discussion
Current Student B
Joined: 29 Mar 2012
Posts: 303
Location: India
GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3: 730 Q50 V38 Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

1
1
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: 137
Location: New Delhi, India
Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

1
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
Current Student B
Joined: 29 Mar 2012
Posts: 303
Location: India
GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3: 730 Q50 V38 Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

1
1
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 V
Joined: 02 Sep 2009
Posts: 53831
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

2
1
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: 406
Location: India
GPA: 2.5
WE: Operations (Hospitality and Tourism)
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

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+++

-----------------------------
Quant Notes consolidated: http://gmatclub.com/forum/consolodited-quant-guides-of-forum-most-helpful-in-preps-151067.html#p1217652

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

Math Expert V
Joined: 02 Sep 2009
Posts: 53831
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

1
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.
_________________
Intern  B
Joined: 28 Jun 2011
Posts: 12
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 53831
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

1
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: 1817
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

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

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 53831
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

1
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!

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
_________________
Intern  Joined: 29 Oct 2014
Posts: 24
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

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: 1817
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

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: 22 Oct 2014
Posts: 27
If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

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

### Show Tags

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.
Senior Manager  B
Joined: 10 Mar 2013
Posts: 496
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24 GPA: 3.88
WE: Information Technology (Consulting)
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

see screenshot below
Attachments Math.jpg [ 51.45 KiB | Viewed 25472 times ]

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

Share some Kudos, if my posts help you. Thank you !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660
e-GMAT Representative D
Joined: 04 Jan 2015
Posts: 2722
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

2
melaos wrote:
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.

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
_________________  Number Properties:- Get 5 free video lessons, 50 practice questions | Algebra:-Get 4 free video lessons, 40 practice questions
Quant Workshop:- Get 100 practice questions | Free Strategy Session:- Key strategy to score 760

Success Stories
Q38 to Q50- Interview call form Wharton | Q35 to Q50 | More Success Stories

Ace GMAT
Articles and Questions to reach Q51 | Question of the week | Tips From V40+ Scorers | V27 to V42:- GMAT 770 | Guide to Get into ISB-MBA

Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Retired Moderator Joined: 29 Apr 2015
Posts: 838
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

### Show Tags

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

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
_________________
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. Re: If root{3-2x} = root(2x) +1, then 4x^2 =   [#permalink] 30 May 2015, 04:33

Go to page    1   2    Next  [ 36 posts ]

Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.  