It is currently 19 Oct 2017, 13:49

### 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

# (1+ (3)^1/3) (2+(3^1/2))^1/2?

Author Message
TAGS:

### Hide Tags

Intern
Joined: 18 Apr 2013
Posts: 27

Kudos [?]: 37 [0], given: 3

### Show Tags

06 Jul 2017, 07:18
11
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

49% (02:37) correct 51% (02:03) wrong based on 106 sessions

### HideShow timer Statistics

$$(1+ \sqrt{3})*(\sqrt{2+\sqrt{3}})$$

A. $$\sqrt{2}(2+\sqrt{3})$$

B. $$\sqrt{2}(3+\sqrt{3})$$

C. $$\sqrt{2}(2-\sqrt{3})$$

D. $$\sqrt{3}(2+\sqrt{3})$$

E. $$\sqrt{2}(2+\sqrt{2})$$

[Reveal] Spoiler:
Attachment:

Capture.JPG [ 17.72 KiB | Viewed 962 times ]
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Oct 2017, 02:16, edited 2 times in total.
Formatted the question.

Kudos [?]: 37 [0], given: 3

BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 1454

Kudos [?]: 594 [3], given: 16

Location: India
WE: Sales (Retail)

### Show Tags

06 Jul 2017, 08:20
3
KUDOS
1
This post was
BOOKMARKED
Let X = $$(1+ \sqrt{3})\sqrt{(2+3^{1/2})}$$ = $$(1+ \sqrt{3})\sqrt{(1 + 1 + 3^{1/2})}$$

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

The expression now becomes X = $$a * \sqrt{1 + a}$$

Squaring
$$X^2 = a^2(1+a)$$

Substituting values,
$$X^2 = (4 + 2\sqrt{3})(2 + \sqrt{3})$$
$$= 2(2 + \sqrt{3})(2 + \sqrt{3})$$

Taking square root
X = $$\sqrt{2}(2 + \sqrt{3})$$ (Option A)
_________________

Stay hungry, Stay foolish

Kudos [?]: 594 [3], given: 16

Manager
Joined: 09 Dec 2015
Posts: 113

Kudos [?]: 18 [3], given: 48

Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q49 V36
GPA: 3.5
WE: Engineering (Manufacturing)

### Show Tags

10 Jul 2017, 09:29
3
KUDOS
Easy method,
Let (1+√3)√(2+√3) = x. Square both sides
x^2 = (1+√3)^2 * (2+√3)
= (1 + 3 + 2√3) * (2 + √3)
= (4 + 2√3) * (2 + √3)
= 2(2 + √3) * (2 + √3)
= 2 * (2 + √3)^2.
Take square root of both sides, x = √2 * (2 + √3)

Kudos [?]: 18 [3], given: 48

Manager
Joined: 04 May 2014
Posts: 128

Kudos [?]: 9 [0], given: 69

Location: India
WE: Sales (Mutual Funds and Brokerage)

### Show Tags

27 Sep 2017, 09:10
pushpitkc wrote:
Let X = $$(1+ \sqrt{3})\sqrt{(2+3^{1/2})}$$ = $$(1+ \sqrt{3})\sqrt{(1 + 1 + 3^{1/2})}$$

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

The expression now becomes X = $$a * \sqrt{1 + a}$$

Squaring
$$X^2 = a^2(1+a)$$

Substituting values,
$$X^2 = (4 + 2\sqrt{3})(2 + \sqrt{3})$$
$$= 2(2 + \sqrt{3})(2 + \sqrt{3})$$

Taking square root
X = $$\sqrt{2}(2 + \sqrt{3})$$ (Option A)

Are we allowed to take 2 common from one bracket set?

eg 6X7=2(3)(7)

Kudos [?]: 9 [0], given: 69

Manager
Joined: 09 Dec 2015
Posts: 113

Kudos [?]: 18 [0], given: 48

Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q49 V36
GPA: 3.5
WE: Engineering (Manufacturing)

### Show Tags

03 Oct 2017, 00:16
gps5441 wrote:
pushpitkc wrote:
Let X = $$(1+ \sqrt{3})\sqrt{(2+3^{1/2})}$$ = $$(1+ \sqrt{3})\sqrt{(1 + 1 + 3^{1/2})}$$

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

The expression now becomes X = $$a * \sqrt{1 + a}$$

Squaring
$$X^2 = a^2(1+a)$$

Substituting values,
$$X^2 = (4 + 2\sqrt{3})(2 + \sqrt{3})$$
$$= 2(2 + \sqrt{3})(2 + \sqrt{3})$$

Taking square root
X = $$\sqrt{2}(2 + \sqrt{3})$$ (Option A)

Are we allowed to take 2 common from one bracket set?

eg 6X7=2(3)(7)

Yes you can take '2' common from one braket.

Notice that when you multiply '2' back to the braket, you'll get the previous equation.

I'll explain with example.

Case 1: Let us say an equation is x = (2y + 4)*(y^2 + 3y); here x and y are some variables.
In this case you can take '2' common from first bracket and 'y' common from second bracket. Equation can be written as, x = 2y*(y+2)*(y+3).
You can also multiply '2' and 'y' separately to the brackets, even to the bracket from which each was not taken.
Equation can also be written as, x = (y^2+2y)*(2y+6) = 2*(y^2+2y)*(y+3) = y*(y+2)*(2y+6). All these equations are same and correct.

Case 2: Let us say an equation is x = (2y + 4) + (y^2 + 3y); here x and y are some variables.
In this case also you can take '2' common from first bracket and 'y' common from second bracket but you cannot multiply both commons like in previous case.
Equation will be written as, x = 2(y+2) + y(y+3). This equation is same as original equation.

Case 3: Let us say an equation is x = (3y + 6) + (y^2 + 2y); here x and y are some variables.
In this case also you can take '2' common from first bracket and 'y' common from second bracket. Equation can be written as, x = 3(y+2)+y(y+2).
In this case, you can take '(y+2)' as common and equation will become, x = (y+2)*(y+3).

I hope this clears your doubt.

Kudos [?]: 18 [0], given: 48

Intern
Joined: 31 Aug 2017
Posts: 1

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

### Show Tags

03 Oct 2017, 02:15

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

Re: (1+ (3)^1/3) (2+(3^1/2))^1/2?   [#permalink] 03 Oct 2017, 02:15
Display posts from previous: Sort by