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

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Intern
Joined: 18 Apr 2013
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06 Jul 2017, 06:18
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50% (02:22) correct 50% (02:14) wrong based on 137 sessions

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

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:
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[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Oct 2017, 01:16, edited 2 times in total.
Formatted the question.
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06 Jul 2017, 07:20
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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)
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Manager
Joined: 09 Dec 2015
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Location: India
Concentration: General Management, Operations
Schools: IIMC (I)
GMAT 1: 700 Q49 V36
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WE: Engineering (Consumer Products)

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10 Jul 2017, 08:29
4
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)
Manager
Joined: 03 May 2014
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27 Sep 2017, 08:10
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)
Manager
Joined: 09 Dec 2015
Posts: 123
Location: India
Concentration: General Management, Operations
Schools: IIMC (I)
GMAT 1: 700 Q49 V36
GPA: 3.5
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02 Oct 2017, 23:16
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.
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03 Oct 2017, 01:15
Re: (1+ (3)^1/3) (2+(3^1/2))^1/2?   [#permalink] 03 Oct 2017, 01:15
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