GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 27 Jan 2020, 04:30 ### 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.  # (1+ (3)^1/3) (2+(3^1/2))^1/2?

Author Message
TAGS:

### Hide Tags

Intern  B
Joined: 18 Apr 2013
Posts: 33

### Show Tags

17 00:00

Difficulty:   85% (hard)

Question Stats: 49% (02:22) correct 51% (02:39) wrong based on 185 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. $$\sqrt{2}(2+\sqrt{2})$$

Attachment: Capture.JPG [ 17.72 KiB | Viewed 2504 times ]

Originally posted by roastedchips on 06 Jul 2017, 07:18.
Last edited by Bunuel on 03 Oct 2017, 02:16, edited 2 times in total.
Formatted the question.
Current Student B
Joined: 09 Dec 2015
Posts: 110
Location: India
Concentration: General Management, Operations
Schools: IIMC (A)
GMAT 1: 700 Q49 V36 GPA: 3.5
WE: Engineering (Consumer Products)

### Show Tags

8
1
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)
##### General Discussion
Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3284
Location: India
GPA: 3.12

### Show Tags

3
2
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, we get $$X^2 = a^2(1+a)$$

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

Taking square root, we have X = $$\sqrt{2}(2 + \sqrt{3})$$ (Option A)
_________________
You've got what it takes, but it will take everything you've got
Manager  B
Joined: 04 May 2014
Posts: 147
Location: India
WE: Sales (Mutual Funds and Brokerage)

### Show Tags

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)
Current Student B
Joined: 09 Dec 2015
Posts: 110
Location: India
Concentration: General Management, Operations
Schools: IIMC (A)
GMAT 1: 700 Q49 V36 GPA: 3.5
WE: Engineering (Consumer Products)

### Show Tags

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.
Intern  Joined: 31 Aug 2017
Posts: 1

### Show Tags

Manager  S
Joined: 28 Jan 2019
Posts: 96
Location: India
GMAT 1: 700 Q49 V36 GPA: 4
WE: Manufacturing and Production (Manufacturing)

### Show Tags

Hey can someone help me out here pls.. here's where I'm stuck.

Squaring the expression = (1+3+2√3)(2+√3)
=2+√3+6+3√3+4√3+6
=14+8√3 (..At this point I'm pretty sure that Im going the wrong way.)
=6+8+8√3
=6+8(1+√3)
=??
_________________
"Luck is when preparation meets opportunity!"
Manager  B
Joined: 10 May 2018
Posts: 75

### Show Tags

1
OhsostudiousMJ

Squaring the expression = (1+3+2√3)(2+√3)
=2+√3+6+3√3+4√3+6
=14+8√3 (..At this point I'm pretty sure that Im going the wrong way.)
=6+8+8√3
=6+8(1+√3

==> 14 +8√3
=8 +6+4√3 +4√3
=4(2+√3) +2√3(2+√3)
=(4+2√3)(2+√3)
=2(2+√3) (2+√3)
=2(2+√3)^2
x =√2(2+√3) (1+ (3)^1/3) (2+(3^1/2))^1/2?   [#permalink] 02 Mar 2019, 13:17
Display posts from previous: Sort by

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